Category: CBSE Class 10

RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself are helpful to complete your math homework.

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RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Correct option: (d)

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

Question 1.
Solution:
Let the length of plot be x meters
Its perimeter = 2 [length + breadth]
= 2(x + 16) = (2x + 32) meters

Length of the rectangle is 21. 5 meter
Area of the rectangular plot = length × breadth
= (16 × 21.5)m²
= 344 m²
The length = 21.5 m and the area = 344 m²

Question 2.
Solution:
The length of a rectangular park is twice its breadth
and its perimeter is 840 m.
Find the area of the park.
Let the breadth of a rectangular park = x m
Then, length of a rectangular park = 2x m
Perimeter of a rectangular park = 840 m

∴ Area of a rectangular park = length x breadth = 140x 280 = 39200 m2

Question 3.
Solution:
Let ABCD be the rectangle in which AB = 12 cm and AC = 37 m

Thus, length = 35 cm and breadth = 12 cm
Area of rectangle = (12 × 35) = 420
Hence, the other side = 35 cm and the area = 420 cm²

Question 4.
Solution:
Let the breadth of the plot be x meter
Area = Length × Breadth = (28 × x) meter
= 28x m²
Breadth of plot is = 16. 5 m
Perimeter of the plot is = 2(length + breadth)

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:
Area of verandah = (36 × 15)m² = 540 m²
Area of stone = (0.6 × 0.5)m² [10 dm = 1 m]
Number of stones required = \(\frac { Area\quad of\quad verandah }{ Area\quad of\quad stone } =\frac { 540 }{ 0.3 } =1800\)
Hence, 1800 stones are required to pave the verandah

Question 9.
Solution:

Question 10.
Solution:
Length of the park = 35 m
Breadth of the park = 18 m

Area of the park = (35 18)m² = 630 m²
Length of the park with grass =(35 – 5) = 30 m
Breadth of the park with grass = (18 – 5) m = 13 m
Area of park with grass = (30 × 13) m² = 390 m²
Area of path without grass = Area of the whole park area of park with grass
= 630 m² – 390 m² = 240 m²
Hence, area of the park to be laid with grass = 240 m²

Question 11.
Solution:
Length of the plot = 125 m
Breadth of the plot = 78 m

Area of plot ABCD = (125 × 78)m² = 9750 m²
Length of the plot including the path= (125 + 3 + 3) m = 131 m
Breadth of the plot including the path = (78 + 3 + 3) m = 84 m
Area of plot PQRS including the path
= (131 × 84) m² = 11004 m²
Area of path = Area of plot PQRS Area of plot ABCD
= (11004 – 9750)m² = 1254 m²
Cost of gravelling = Rs 75 per m²
Cost of gravelling the whole path = Rs. (1254 × 75) = Rs. 94050
Hence, cost of gravelling the path = Rs. 94050.

Question 12.
Solution:
Area of rectangular field including the foot path = (54 × 35)m²
Let the width of the path be x m
Then, area of rectangle plot excluding the path = (54 × 2x) (35 × 2x)
Area of path = (54 × 35) (54 × 2x) (35 × 2x) = 54 ×35) (54 × 2x) (35 × 2x) = 420

Question 13.
Solution:
Let the length and breadth of a rectangular garden be 9x and 5x.

Then, area of garden = (9x 5x)m = 45x² m²
Length of park excluding the path = (9x 7) m
Breadth of the park excluding the path = (5x 7) m
Area of the park excluding the path = (9x 7)(5x 7) m²

Question 14.
Solution:

Question 15.
Solution:
Let the width of the carpet = x meter
Area of floor ABCD = (8 ×5) m²
Area of floor PQRS without border
= (8 × 2x)(5 × 2x)
= 40 × 16 x 10x + 4x²
= 40 × 26x + 4x²
Area of border = Area of floor ABCD Area of floor PQRS

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

Question 28.
Solution:
Area of the ||gm = (base height) sq. unit
= (25 × 16.8) cm² = 420 cm²

Question 29.
Solution:

Question 30.
Solution:

Question 31.
Solution:
Area of parallelogram = 2 area of DABC
Opposite sides of parallelogram are equal
AD = BC = 20 cm
And AB = DC = 34 cm
In ∆ABC we have
a = AC = 42 cm
b = AB = 34 cm
c = BC = 20 cm

Question 32.
Solution:

We know that the diagonals of a rhombus, bisect each other at right angles
OA = OC = 15 cm,
And OB = OD = 8 cm
And ∠AOB = 90⁰
∴By Pythagoras theorem, we have

Question 33.
Solution:

Question 34.
Solution:

Question 35.
Solution:

Question 36.
Solution:

Question 37.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17a

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17a.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

Question 1.
Solution:

Question 2.
Solution:
Let a = 42 cm, b = 34 cm and c = 20 cm

(ii)Let base = 42 cm and corresponding height = h cm

Hence, the height corresponding to the longest side = 16 cm

Question 3.
Solution:
Let a = 18 cm, b = 24 cm, c = 30 cm
Then,2s = (18 + 24 + 30) cm = 72 cm
s = 36 cm
(s a) = 18cm, (s b) = 12 cm and (s c) = 6 cm

(ii)Let base = 18 cm and altitude = x cm

Hence, altitude corresponding to the smallest side = 24 cm

Question 4.
Solution:
On dividing 150 m in the ratio 5 : 12 : 13, we get

Question 5.
Solution:
On dividing 540 m in ratio 25 : 17 : 12, we get

Question 6.
Solution:
Let the length of one side be x cm
Then the length of other side = {40 (17 + x)} cm = (23 – x) cm
Hypotenuse = 17 cm
Applying Pythagoras theorem, we get

Question 7.
Solution:
Let the sides containing the right – angle be x cm and (x – 7) cm

∴ perimeter of triangle (15 + 8 + 17) cm = 40 cm

Question 8.
Solution:
Let the sides containing the right angle be x and (x 2) cm

Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cm

Question 9.
Solution:
Side of an equilateral triangle = a = 10 cm

Question 10.
Solution:
Let each side of the equilateral triangle be a cm

Question 11.
Solution:
Let each side of the equilateral triangle be a cm

Perimeter of equilateral triangle = 3a = (3 12) cm = 36 cm

Question 12.
Solution:
Let each side of the equilateral triangle be a cm

Question 13.
Solution:
Base of right angled triangle = 48 cm

Question 14.
Solution:
Let the hypotenuse of right – angle triangle = 6.5 m
Base = 6 cm

Hence, perpendicular = 2.5 cm and area of the triangle =7.5 cm²

Question 15.
Solution:
The circumcentre of a right – triangle is the midpoint of the hypotenuse

Hypotenuse = 2 × (radius of circumcircle)
= (2 × 8) cm = 16 cm
Base = 16 cm, height = 6 cm
Area of right angled triangle=

Hence, area of the triangle= 48 cm²

Question 16.
Solution:
Let each equal side be a cm in length.
Then,

Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm

Question 17.
Solution:
Let each equal side be a cm and base = 80 cm

perimeter of triangle = (2a + b) cm
= (2 41 + 80) cm
= (82 + 80) cm = 162 cm
Hence, perimeter of the triangle = 162 cm

Question 18.
Solution:
Let the height be h cm, then a= (h + 2) cm and b = 12 cm

Hence, area of the triangle = 48 cm².

Question 19.
Solution:
Let ∆ABC is a isosceles triangle. Let AC, BC be the equal sides
Then AC = BC = 10cm. Let AB be the base of ∆ABC right angle at C.

Hence, area = 50 cm² and perimeter = 34.14 cm

Question 20.
Solution:

Area of shaded region = Area of ABC – Area of DBC
First we find area of ABC

Area of shaded region = Area of ∆ABC – Area of ∆DBC
= (43.30 – 24) cm² = 19. 30 cm²
Area of shaded region = 19.3 cm²

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17a are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Practical Based Questions for Class 10 Science Chemistry

Question 1.
A student dipped a strip of pH paper in distilled water taken in a tube. As expected, the pH paper acquired green colour. He then dissolved a pinch of common salt in the same tube. What will be the expected change in colour of the pH paper ?
Answer:
There will not be any change in the colour of the pH paper. It will remain green which indicates that the solution has maintained neutral character. Actually, sodium chloride is a salt prepared from strong base (NaOH) and strong acid (HCl). It is a neutral salt and is not expected to bring any change in the pH of distilled water.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
(a) Five solutions A, B, C, D and E when tested with universal indicator showed pW as 4, 1, 11,7 and 9 respectively. Which solutions is :

1. neutral
2. strongly alkaline
3. strongly acidic
4. weakly acidic acid
5. weakly alkaline.

Arrange the solutions in increasing order of H-ion concentration.
Answer:

1. D
2. C
3. B
4. A
5. E.

The increasing order of H-ion concentration :C<E<D<A<B.

Question 3.
Name the acid and base from which the following salts have been formed :

1. Sodium acetate
2. Ammonium chloride
3. Calcium nitrate
4. Sodium carbonate.

Answer:

1. NaOH and CH₃COOH
2. NH₄OH and HCl
3. Ca(OH)₂ and HNO₃
4. NaOH and H₃CO₃.

Question 4.
When water is added gradually to a white solid X, a hissing sound is heard and a lot of heat is produced forming the product Y. A suspension of Y in water is applied to the walls of a house during white washing.
(a) What could be the solid X ? Write its chemical formula.
(b) What could be the product Y ? Write its chemical formula.
Answer:
(a) The white solid X is quick lime. Its chemical formula is CaO.
(b) When water is added to the white solid X, product Y formed is calcium hydroxide. Its chemical formula is Ca(OH)r

Question 5.
Two drinks P and Q gave acidic and alkaline reactions respectively. One has a pH value of 9 and the other has pH value of 3. Which drink has the pH value of 9 ?
Answer:
The solution Q is of alkaline nature. It has a pH value of 9.

Question 6.
Take five test tubes and label them as A, B, C, D and E. Add 5 mL of five unknown solutions in them. Put a small strip of universal indicator in each of them. Following colours appear in these :
Solution A — Orange;
Solution B — Green;
Solution  C – Red;
Solution D – Blue;
Solution E – Violet.
Predict the nature of the solutions in these from the pH character.
Answer:
Solution-A (Weakly acidic);
Solution-B (Neutral);
Solution-C (Strongly acidic);
Solution-D (Weakly basic);
Solution-E (Strongly basic).

Question 7.
A metal A’ gives a compound ‘B’ (molecular mass 40) when it reacts with water. Compound ‘B’ gives a soluble compound ‘C’ on treatment with aluminium oxide. Identify A’, ‘B’ and ‘C’. Also give the reactions involved.
(CBSE 2013)
Answer:
The available data suggests that the metal A sodium (Na). It reacts with water to form sodium hydroxide ‘B’ with formula NaOH. On reacting with aluminium oxide, the compound ‘B’ forms sodium metaaluminate ‘C.
The chemical equations for the reactions are :

Question 8.
You want to study a decomposition reaction by taking ferrous sulphate crystals in a boiling tube. List two steps you would follow while doing the experiment. (CBSE 2014)
Answer:

1. Take a test tube and dry it completely.
2. Take a small amount of the given sample in the test tube. Hold it with a clamp and heat the tube over a burner. Crystals will first become dirty white and then change to brown.

Question 9.
On keeping iron nails in a blue coloured copper sulphate solution, it is observed that the colour of the solution turns light green after sometime. Give reasons for this colour change. Name the type of reaction.
(CBSE 2014)
Answer:
It is an example of metal displacement reaction. Iron has displaced copper from copper sulphate solution and has changed to iron sulphate which is light green in colour.

Question 10.
While studying the double displacement reaction, the solutions of barium chloride and sodium sulphate are mixed together.
(i) What do you observe as soon as the two solutions are mixed together ?
(ii) What will happen in the above observation made by you after ten minutes ? (CBSE 2015)
Answer:
(i) A white precipitate of barium sulphate is immediately formed.

(ii) The white precipitate will settle down at the bottom of the tube and the solution above the precipitate will become colourless.

Question 11.
You want to perform an experiment to study a double displacement reaction in your school laboratory. Name two aqueous solutions required for the experiment. State the colour change you are likly to observe on mixing the two solutions. (CBSE 2015)
Answer:
(i) A white precipitate of barium sulphate is immediately formed.

(ii) The white precipitate will settle down at the bottom of the tube and the solution above the precipitate will become colourless.

Question 12.
You are given two colourless solutions present in two test tubes. One out of these is ethyl alcohol and the other is acetic acid. Give three tests to identify these.
Answer:
Smell. The tube in which the liquid has a vinegar like or fruity smell is acetic acid while the other with a very little pungent smell is ethyl alcohol.
Litmus Test. Add a strip of blue litmus separately in both the tubes. The tube which turns blue litmus red contains acetic acid while the other which does not change its colour is that of ethyl alcohol.
Sodium hydrogen carbonate test. Add a small amount of solid sodium hydrogen carbonate (NaHCO₃) in both these tubes. The one which gives a brisk effervescence contains acetic acid while the other in which no effervescence is noticed contains ethyl alcohol.

Question 13.
You are provided with two samples of hard water; one containing temporary hardness and the other permanent hardness. Without the help of any chemical, how will you identify the nature of the sample.
Answer:
Boil the two samples separately in beakers for sometime. If precipitate appears, filter it out. Now, add a strip of soap in each. The one which produces lather contained temporary hardness while the other in which no lather is formed, contained permanent hardness.

Question 14.
How will you distinguish between ethane and ethene with the help of a chemical test ?
Answer:
Pass the two gases separately through bromine dissolved in carbon tetrachloride taken in two test tubes. If the yellow colour of bromine gets discharged, the gas is ethene. If the colour remains intact, then the gas is ethane.

Question 15.
An unknown organic liquid does not turn blue litmus red and gives no effervescence with sodium hydrogen carbonate. However, when a dry piece of sodium pellet is added to the liquid, a gas is evolved with brisk effervescence. Identify the liquid.
Answer:
The given liquid is probably an alcohol (e.g., ethyl alcohol) which has no reaction with blue litmus or sodium hydrogen carbonate. But it evolves hydrogen gas on reacting with sodium metal
2C₂H₅OH + 2Na ————> 2C₂H₅ONa + H₂

Question 16.
How will you distinguish between hydrochloric acid and ethanoic acid with the strip of a-universal pH paper ?
Answer:
The strip of universal pH paper will turn red in hydrochloric acid while its colour will change to orange in ethanoic acid. In fact, hydrochloric acid is a stronger acid than acetic acid and they have different pH values.

Question 17.
Give a simple test to distinguish soaps from detergents.
Answer:
Soaps donot give lather with hard water. However, detergents form lather with hard water.

Question 18.
Write the name of apparatus/chemicals required to study the following properties of ethanoic acid in the laboratory.
Nature, odour, solubility and action are sodium hydrogen carbonate. (CBSE 2015)
Answer:
Litmus paper, test tube, test tube, lime water.

Question 19.
A student added sodium hydrogen carbonate solution in ethenoic acid taken in a test tube and the gas evolved was tested with a burning splinter. Write the chemical equation for the evolution of this gas and its effect on burning splinter. (CBSE 2015, Sample Paper 2017)
Answer:
CH₃COOH + NaHCO₃ ———–> CH₃COONa + H₂O + CO₂
Since CO₂ is not a supporter of combustion, the burning splinter will be extinguished.

Question 20.
A student is studying the properties of acetic acid in his school laboratory. List two physical and two chemical properties which he must observe and note in his record book. (CBSE 2016)
Answer:
Complete oxidation: When ethanol is warmed with dilute alkaline solution of potassium permanganate (5% solution) called Baeyer’s reagent, ethanoic acid or acetic acid is formed as the product.
CH₃CH₂OH + 2(O) ————–> CH₃COOH +H₂O
Partial oxidation: Upon oxidation with chromic anhydride (CrO₃) dissolved in acetic acid (CH₃COOH), ethanol forms ethanal, also called acetaldehyde.
CH₃CH₂OH +  (O)  —————– > CH₃CHO + H₂O

Question 21.
A student adds a spoon full of powdered sodium hydrogen carbonate to a flask containing ethanoic acid. List two main observations, he must note in his note book, about the reaction that takes place. Also write chemical equation for the reaction. (CBSE 2016)
Answer:

1. A colourless and odourless gas is evolved accompanied by brisk effervescence.
2. When the gas is bubbled through lime water, it becomes milky.

Question 22.
A gas is liberated immediately with brisk effervescence when you add acetic acid to sodium hydrogen carbonate powder in a test tube. Name the gas and describe a test to confirm the identity of the gas.
(CBSE 2017)
Answer:
The gas evolved is CO₂
CH₃COOH + NaHCO₃ ———–> CH₃COONa + H₂O + CO₂
It turns lime water milky when bubbled in small amount. On bubbling the gas in excess, the milkiness disappears.

Question 23.
If you are asked to report two observations about the following two properties of acetic acid, what would you report (t) Odour (it) Effect on litmus. (CBSE 2017)
Answer:

1. Acetic acid has vinegar smell.
2. On adding a few drops of blue litmus solution to acetic acid, it acquires red colour.

Question 24.
Mention the essential materials (chemicals) to prepare soap in the laboratory. Describe in brief the test of determining the nature (acidic/alkaline) of the reaction mixture of a saponification reaction.
(CBSE 2017)
Answer:
The essential materials are edible oil (from animal or vegetable origin), caustic alkali (NaOH or KOH) and common salt. The reaction mixture in the saponification reaction is of alkaline nature since it contains in it a caustic alkali. The nature can be tested by adding a few drops of red litmus to the mixture. Its colour will change to blue.

Question 25.
What do you observe when you add a few drops of acetic acid to a test tube containing.
(a) phenolphthalein
(b) distilled water
(c) universal indicator
(d) sodium hydrogen carbonate.
(CBSE 2017)
Answer:
(a) Acetic acid will remain colourless in phenolphthalein.
(b) Acetic acid will mix with distilled water to form a clear solution.
(c) A universal indicator will impart orange colour to acetic acid.
(d) A brisk effervescence will be noticed due to the evolution of CO₂ gas.

 

Hope given Practical Based Questions for Class 10 Science Chemistry helpful to you.

If you have any doubts, please comment below. We try to provide online math tutoring for you.

Value Based Questions in Science for Class 10 Chapter 4 Carbon and Its Compounds

Question 1.
Kamla and Reema are best friends. On one evening Kamla went to the house of Reema and found her working in the kitchen. The gas burner was emitting yellow flame instead of blue flame. Kamla immediately asked Reema to put off the gas. She helped her in cleaning the fine holes of the gas burner with a needle. The entire operation took about fifteen minutes. The gas was now ignited and there was a blue flame. Please read the above narration and answer the following questions :

1. Why was burner emitting yellow flame ?
2. What was the purpose of cleaning the holes ?
3. In what way Kamla helped Reema ?
4. What lesson can we learn from this ?

Answer:

1. The holes of the burner were blocked due to soot or some oily material released during cooking. The combustion was incomplete and the gas burnt with yellow flame.
2. By cleaning the holes, the deposits were removed and the combustion was now complete.
3. By doing this operation, there was a saving of fuel. This also helped in checking pollution.
4. We should always keep gas burners and engines of scooters or cars clean so that the combustion is complete and there is no wastage of any fuel. This also checks pollution problem.

More Resources

• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Mohan and Sohan were working in a factory and were good friends. On one evening Sohan received a phone . call from Mohan that his eye sight had become very dim all of a sudden. Sohan immediately rushed to his place and came to know that this had occurred after consuming some alcohol. Sohan took him to an eye specialist. He checked his vision properly and gave some antibiotic drops. The specialist kept him under observation and by next morning, he was fully cured.

1. Why did eye sight of Mohan become dim ?
2. How was he cured ?
3. What is the value associated with episode ?

Answer:

1. Mohan had consumed adultrated liquor poisoned either with methyl alcohol or copper sulphate.
2. Antibiotic given by the specialist neutralised the effect of poisoning. As a result, Mohan regained his sight.
3. Timely help by Sohan saved Mohan from getting completely blind. This was need a service rendered by one friend to the other.

Question 3.
Teacher asked Hema to perform test for unsaturation in the laboratory for ethylene gas. She took some chlorine water in a tube and passed the vapours of the gas. Nothing happened Teacher asked her to pass the vapours of the gas into bromine water. The yellow colour of the gas immediately discharged.

1. What was the mistake committed by Hema ?
2. How did teacher help her ?
3. Write chemical equation for the reaction.

Answer:

1. Chlorine water has no colour. Therefore, on passing ethylene gas the colour of chlorine water did not discharge.
2. Bromine water is yellow in colour. When ethylene gas was passed through bromine water, its colour got discharged. This is the test for unsaturation.
3. The chemical equation for the reaction is :
   

Question 4.
A patient was suffering from high blood pressure and high cholesterol.He went to the specialist. He enquired about his eating habits. The patient told the doctor that he consumes both dalda ghee and desi ghee and also drinks milk with full fat. The doctor asked the patient to immediately stop these and instead use vegetable oil and also drink fat free milk.

1. What was wrong with the eating habits of the patient ?
2. How did doctor help him ?

Answer:

1. Both dalda/desi ghee and milk full of fats are very harmful to our body. The fats get deposited and lead to high chrolestrol level and high blood pressure. This may ultimately result in either paralysis or death.
2. Doctor gave the patient correct advice. The diet should be free from fats as much as possible particularly in the present set up when there is lack of exercise.

Question 5.
Ethanol, commonly called alcohol is an excellent solvent, is used in medicines and is an important chemical compound involved in synthesis of many chemical compounds. However in spite of its benefits to man, its impact on social behaviour has always been questioned. Media has often shown abnormal behaviour of people while drunk. It is considered as a curse in the lives of those who are addicted to alcohol – Alcoholic’ people are not only lowering their metabolism and affecting Central Nervous System, they are also a threat to the lives of others. Anger and rude behaviour are some of its ill effects.

1. Comment on the statement – ‘Should production of alcohol be banned’ give three valid reasons to justify.
2. As a student what initiative would you take in the common concern of‘Save Life, Do not Drink’. Give two suggestions.

Answer:
In favour of negative response :

1. Ethanol is used as a solvent in the laboratory.
2. Ethanol is used in the synthesis of large number of compounds.
3. Ethanol is a constituent of many drugs.

In favour of positive response :

1. Ethanol is habit forming.
2. Excessive intake of ethanol is highly injurious to the body.
3. Excessive consumption of ethanol is the major cause of crimes in our society.
4. It is a big strain on the family budget. Many families are ruined.

Initiatives :

1. Students must take a pledge not to drink alcohol.
2. Students must create awareness in the society by organising skits/seminars and also through charts.

Question 6.
School going children generally bring tiffins and they eat food during the break. One category of students marked ‘A’ carry paranthas, butter and pickels while the other category of students marked ‘B’ bring chapatties, vegetables, salad and fruits .Whereas students A’ donot like to share their food, students ‘B’ would like to share it.

1. Which acid is present in pickels ?
2. Which group of students bring healthy food and why ?
3. Which group of students bring unhealthy food and why ?
4. Which group of students have better value system and why ?

Answer:

1. Acetic acid is present in pickels
2. Students marked ‘B’ bring healthy food since it has more neutrition values and it is easy to digest.
3. Students marked A’ bring unhealthy food. Since it is quite difficulty to digest it and leads to obacity.
4. Students marked ‘B’ have better value system. Sharing of food at school level will make them better citizens at a later stage in their life.

Question 7.
Cough syrups, generally contain alcohol. Some people are habitual of drinking ‘alcohol’. Instead of drinking ‘alcohol’, they have started using cough syrups which contains alcohol and cause addiction. To solve this problem, government is thinking to ban cough syrups.

1. What is an alcohol ?
2. Should production of cough syrups be banned ?
3. As a student ‘what initiative would you take to make people aware of harmful effects of taking cough syrups unnecessarily’. Give two suggestions.

Answer:

1. Alcohol is chemically ethyl alcohol and its formula is C₂H₅OH.
2. No, production of all cough syrups cannot be banned since they given relief from cough and cold. However, they should be sold strictly if prescribed by the doctor on a proper prescription.
3. We students must acquaint people around us the harmful effects of alcohols by holding seminars and by door to door convassing.

 

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Value Based Questions in Science for Class 10 Chapter 3 Metals and Non-metals

These Solutions are part of Value Based Questions in Science for Class 10. Here we have given Value Based Questions in Science for Class 10 Chapter 3 Metals and Non-metals

Question 1.
A customer approached a newly trained Goldsmith and asked him to make some ornaments. For this he gave advance money also. The Goldsmith bought a bar of 24-carat gold from the market and tried to make ornaments from it. He was not successful because the gold was too soft to handle. He approached a trained Goldsmith who asked him to use 22-carat gold for this purpose. He was now quite successful and could make the ornaments ?

1. What is the difference between 24-carat gold and 22-carat gold ?
2. What was wrong with 24-carat gold ?
3. How did trained Goldsmith help the freshly trained Goldsmith ?
4. What is the value associated with this episode ?

Answer:

1. 24-carat gold is pure gold while 22-carat gold is comparatively less pure.
2. Pure or 24-carat gold is very soft, little malleable and ductile. It is quite difficult to work on it. But 22 carat gold (alloy) is comparatively hard, more malleable and ductile
3. Freshly trained Goldsmith realised his mistake and could complete the order. Otherwise he might have been in problem.
4. The trained Goldsmith had a big heart and did not bother about the professional jealousy. He was very sincere in his approach and helped the junior colleage.

More Resources

• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
A student was asked to purify a sample of copper extracted by a suitable method. He constructed a-cell in which a rod of impure copper was made cathode while that of pure copper as anode. The electrolyte was aqueous CuSO₄ solution. On passing electric current nothing happened. He realised his mistake and connected the electrodes in the reverse order. He was now quite successful in his mission.

1. What was his mistake ?
2. How did he rectify it ?
3. Write a chemical equation for the reaction.

Answer:

1. Impure copper cannot be made cathode and pure copper as the anode.
2. He changed over. Pure copper was made cathode while impure copper as the anode.
3. The chemical equations for the process are :
   

 

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HOTS Questions for Class 10 Science Chapter 3 Metals and Non-metals

These Solutions are part of HOTS Questions for Class 10 Science. Here we have given HOTS Questions for Class 10 Science Chapter 3 Metals and Non-metals

Question 1.
A student has been collecting silver coins and copper coins. One day she observed a black coating on silver coins and a green coating on copper coins. Which chemical phenomenon is responsible for these coatings ? Write the chemical names of black and green coatings ?
Answer:
The phenomenon is known as corrosion. Air contains traces of hydrogen sulphide gas which reacts with silver metal present in the coin to form silver sulphide. It is black in colour.

Similarly, copper present in the coin reacts with oxygen and traces of both carbon dioxide and water vapours present in air to form a green mass. It is chemically basic copper carbonate :

More Resources

• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
You are provided with three metals : Sodium, magnesium and copper. Using only water as the reactant, how will you identify them ?
Answer:
The metal which reacts violently with cold water and catches fire is sodium.

1. The metal which evolves hydrogen gas upon heating with water is magnesium.
2. The metal which does not react with water even on strong heating is copper.

Question 3.
E is an element amongst copper, zinc, aluminium and iron. It shows following properties :
(a) One of its ores is rich in E₂O₃
(b) E₂O₃ is not attacked by water,
(c) It forms two chlorides ECl₂ and ECl₃. Name the element and justify your answer.
Answer:
The clue for the correct answer is the formation of ECl₂ and ECl₃. This shows that the element E has variable valencies of 2 and 3. Out of the elements listed, only iron exists in divalent and trivalent forms.
(a) The ore rich in Fe₂O₃ is haematite.
(b) Haematite (Fe₂O₃) is not attacked by water.
(c) The two chlorides are : iron (II) chloride or FeCl₂ and iron (III) chloride or FeCl₃.

Question 4.
An element reacts with oxygen to form an oxide which dissolves in dilute hydrochloric acid. The oxide formed also turns a solution of red litmus blue. Is the element a metal or non-metal ? Explain with the help of a suitable example.
Answer:
The oxide of the element is basic as it turns red litmus solution blue. This means that the element is a metal (M). Let the metal be sodium (Na). The chemical equations that are involved are given as follows :

Question 5.
An element E combines with oxygen to form an oxide E₂O which is a good conductor of electricity.
Give the following information:

1. How many electrons will be present in the valence shell of the element E ?
2. Write the formula of the compound formed when the element E combines with chlorine.

Answer:

1. From the formula E₂O of the oxide, it is clear that the valency of the element E is one. This means that it has only one electron in the valence shell.
2. We know that chlorine is monovalent. Sine the valency of the element E is a also one, the formula of the chloride of the element is ECl.

Question 6.
An element A’ catches fire in water and burns with golden yellow flame in air. It reacts with another element ‘B’, present in group 17 to give a product ‘C’. An aqueous solution of product ‘C’ on electrolysis gives a compound ‘D’ and liberates hydrogen. Identify A, B, C and D.
Answer:
Since the element A’ catches fire in water and burns with golden yellow flame, it is sodium (Na). The element ‘B’ with atomic number 17 is chlorine (Cl). Both these combine to form sodium chloride (NaCl) which is designated as ‘C’. Upon electrolysis, sodium chloride gives sodium hydroxide (D) and evolves hydrogen along with chlorine.

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HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds

Question 1.
An organic compound A of molecular formula C₂H₄ on reduction gives another compound B of molecular formula C₂H₆. B on reaction with chlorine in the presence of sunlight gives C of molecular formula C₂H₅Cl.
(a) Name the compounds A, B and C.
(b) Write chemical equation for the conversion of A to B and name the type of reaction.
Answer:
The compound A of molecular formula C₂H₄ is an alkene. Upon reduction with hydrogen, it gives ‘B’ of molecular formula C₂H₆. The compound ‘B’ upon chlorinadon gives ‘C’ of molecular formula C₂H₅Cl.

The reaction to called addition reaction.

More Resources

• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Name the functional group of organic compounds that can be hydrogenated. With the help of suitable example, explain the process of hydrogenation mentioning the conditions of the reaction and any one change in physical property with the formation of the product. Name any one natural source of organic compounds that are hydrogenated.
Answer:
The functional group which can be easily hydrogenated is
The family is known as alkenes. The hydrogenation reaction can be carried by heating a member of the family (e.g. ethene) with hydrogen in the presence of a catalyst like nickel (Ni) For example.

Ethene is an unsaturated hydrocarbon while ethane is of saturated nature.
Edible oils such as coconut oil, olive oil, peanut oil etc. contain atleast one
in their molecules. These are regarded as unsaturated compounds. Upon hydrogenation, these get converted into fats which no longer contain any

Question 3.
An organic compound A’ of molecular formula C₂H₆O on oxidation with dilute alkaline KMnO₄ solution gives an acid ‘B’ with the same number of carbon atoms. Compound A’ is often used for sterilization of skin by doctors. Name the compound. Write the chemical equation involved in the formation of ‘B’ from A.
Answer:
The compound ‘B’ should contain a —COOH group as it is an acid. Since it has only two carbon atoms, the other carbon atom must represent CH₃ group. Thus, compound ‘B’ is ethanoic acid (CH₃COOH). The compound A used for the sterilization of skin by doctors is ethanol C₂H₅OH (C₂H₆O). The chemical reaction involved in the oxidation by dilute alkaline KMnO₄ solution also called Baeyers reagent is :

Question 4.
A to F are the structural formulae of some organic compounds :

(i) Give the letters which represent the same family.
(ii) Give the letters which do not represent hydrocarbons.
(iii) Flow can ‘C’ be converted into A ?
Answer:
(i) Letters ‘B’ and ‘D’ represent the family of alkynes.
(ii) Letters ‘E’ and ‘F’ donot represent any hydrocarbon.
(iii) ‘C’ can be converted into ‘A’ by passing hydrogen (H₂) in the presence of Ni at 473 K.

Question 5.
(a) A test tube contains a brown liquid in it. The colour of the liquid remains the same when methane is passed through it but it disappears when ethene is passed. Suggest the name of the liquid brown in colour. Give the chemical equation involved.
(b) The formula of an ester is C₃H₇COOC₂H₅. Write the formulae of the acid and alcohol from which the ester is prepared.
Answer:
(a) The brown liquid seems to be bromine dissolved in water. Methane (CH₄) is a saturated hydrocarbon and does not react with bromine. Ethene (C₂H₄) being unsaturated in nature, decolourises bromine and its colour therefore, disappears.

(b) In an ester, the left side in the molecular formula containing C₃H₇CO is derived from the acid while the right side having OC₂H₅ is from the alcohol. This means that the acid and alcohol participating in the ester are C₃H₇COOH and C₂H₅OH respectively. The formation of ester may be shown as follows :

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NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

NCERT Exemplar Solutions for Class 10 Science Chapter 5 Multiple Choice Questions

Question 1.
Upto which element, the Law of Octaves was found to be applicable ?
(a) Oxygen
(b) Calcium
(c) Cobalt
(d) Potassium.
Answer:
(b). Law was found to be applicable upto the element calcium.

More Resources

• NCERT Exemplar Solutions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
According to Mendeleev’s Periodic Law, the elements were arranged in the periodic table in the order of
(a) increasing atomic number
(b) decreasing atomic number
(c) increasing atomic masses
(d) decreasing atomic masses.
Answer:
(c).

Question 3.
In Mendeleev’s Periodic Table, gaps were left for the elements to be discovered later. Which of the following elements found a place in the periodic table later ?
(a) Germanium
(b) Chlorine
(c) Oxygen
(d) Silicon.
Answer:
(a). The element was called Eka-silicon.

Question 4.
Which of the following statement (s) about the Modern Periodic Table are incorrect ?
(i) The elements in the Modern Periodic Table are arranged on the basis of their decreasing atomic numbers
(ii) The elements in the Modern Periodic Table are arranged on the basis of their increasing atomic masses.
(iii) Isotopes are placed in adjoining group (s) in the Periodic Table
(iv) The elements in the Modern Periodic Table are arranged on the basis of their increasing atomic numbers
(a) (i) only
(b) (i), (ii) and (iii)
(c) (i), (ii) and (iv)
(d) (iv) only.
Answer:
(b). Statements (i), (ii) and (iii) are all in correct.

Question 5.
Which of the following statements about the Modern Periodic Table is correct ?
(a) It has 18 horizontal rows known as Periods
(b) It has 7 vertical columns known as Periods
(c) It has 18 vertical columns known as Groups
(d) It has 7 horizontal rows known as Groups.
Answer:
(c).

Question 6.
Which of the given elements A, B, C, D and E with atomic numbers 2, 3, 7, 10 and 30 respectively belong to the same period ?
(a) A, B, C
(b) B, C, D
(c) A, D, E
(d) B, D, E.
Answer:
(b). The elements B (Z = 3), C (Z = 7) and D(Z = 10) belong to the same period. It is second period.

Question 7.
The elements A, B, C, D and E have atomic numbers 9, 11, 17, 12 and 13 respectively. Which pair of elements belong to the same group ?
(a) A and B
(b) B and D
(c) A and C
(d) D and E.
Answer:
(c). Elements A (Z = 9) and C (Z = 17) belong to the same group. It is a halogen family (group 17).

Question 8.
Where would you locate the element with electronic configuration 2, 8 in the Modern Periodic Table ?
(a) Group 8
(b) Group 2
(c) Group 18
(d) Group 10.
Answer:
(c). It is a noble gas element Neon (Ne) present in group 18.

Question 9.
An element which is an essential constituent of all organic compounds belongs to
(a) group 1
(b) group 14
(c) group 15
(d) group 16.
Answer:
(b). The element is carbon. It belongs to group 14.

Question 10.
Which of the following is the outermost shell for elements of period 2 ?
(a) K shell
(b) L shell
(c) M shell
(d) N shell
Answer:
(b). In second (2) period, the electrons are filled in second shell also known as L-shell.

Question 11.
Which one of the following elements exhibits maximum number of valence electrons ?
(a) Na
(b) Al
(c) Si
(d) P
Answer:
(d). The element phosphorus (P) has five electrons (2, 5) in the valence shell.

Question 12.
Which of the following gives the correct increasing order of the atomic radii of O, F and N ?
(a) O, F, N
(b) N, F, O
(c) O, N, F
(d) F, O, N.
Answer:
(d). It is the correct order. These elements are present in second period.

Question 13.
Which among the following elements has the largest atomic radius ?
(a) Na
(b) Mg
(c) K
(d) Ca.
Answer:
(c). The element potassium (K) present in group 1 has the largest atomic radius.

Question 14.
Which of the following elements would lose an electron easily ?
(a) Mg
(b) Na
(c) K
(d) Ca.
Answer:
(c). The element potassium (K) with maximum size would lose electron easily.

Question 15.
Which of the following elements does not lose an electron easily ?
(a) Na
(b) F
(c) Mg
(d) Al.
Answer:
(b). The element fluorine (F) with smallest size does not lose electron easily.

Question 16.
Which of the following are the characteristics of isotopes of an element ?
(i) Isotopes of an element have same atomic mass.
(ii) Isotopes of an element have same atomic number.
(iii) Isotopes of an element show same physical properties.
(iv) Isotopes of an element show same chemical properties.
(a) (i), (iii) and (iv)
(b) (ii), (iii) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv).
Answer:
(d).

Question 17.
Arrange the following elements in the order of their decreasing metallic character Na, Si, Cl, Mg, Al
(a) Cl > Si > Al > Mg > Na
(b) Na > Mg > Al > Si > Cl
(c) Na > Al > Mg > Cl > Si
(d) Al > Na > Si > Ca > Mg.
Answer:
(b). The metallic character of the elements decreases along a period. These elements are present in third period.

Question 18.
Arrange the following elements in the order of their increasing non-metallic character Li, O, C, Be, F
(a) F < O < C < Be < Li
(b) Li < Be < C < O < F
(c) F < O < C < Be < Li
(d) F < O < Be < C < Li.
Answer:
(b). The non-metallic character of the elements increases along a period. These elements are present in second period in the order Li, Be, C, O, F.

Question 19.
What type of oxide would Eka-aluminium form ?
(a) EO₃
(b) E₃O₂
(c) E₂O₃
(d) EO.
Answer:
(c). Is the correct answer.

Question 20.
Three elements B, Si and Ge are
(a) metals
(b) non-metals
(c) metalloids
(d) metal, non-metal and metalloid respectively.
Answer:
(c). is the correct answer. These are also called semi-metals and possess the characteristics of both metals and non-metals.

Question 21.
Which of the following elements will form an acidic oxide ?
(a) An element with atomic number 7
(b) An element with atomic number 3
(c) An element with atomic number 12
(d) An element with atomic number 19.
Answer:
(a). The element with atomic number (Z) = 7 is nitrogen. It forms acidic oxides such as N₂O₃, N₂O₅ etc.
All other elements are metals and they form basic oxides.

Question 22.
The element with atomic number 14 is hard and forms acidic oxide and a covalent halide. To which of the following categories does the element belong ?
(a) Metal
(b) Metalloid
(c) Non-metal
(d) Left-hand side element.
Answer:
(b). The element with atomic number (Z) = 14 is silicon (Si). It is a metalloid.

Question 23.
Which one of the following depicts the correct representation of atomic radius (r) of an atom ?

(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv).
Answer:
(b). The atom (ii) has only one shell (K-shell). No electrons are present in the other shells. Therefore, the arrow represents correct atomic radius.
The arrow in atom (iii) also represents the correct atomic radius of the element.

Question 24.
Which one of the following does not increase while moving down the group of the periodic table ?
(a) Atomic radius
(b) Metallic character
(c) Valence
(d) Number of shells in an element.
Answer:
(c). In a group, the valence or valency does not change since all the elements present have same valence shell configuration.

Question 25.
On moving from left to right in a period in the periodic table, the size of the atom
(a) increases
(b) decreases
(c) does not change appreciably
(d) first decreases and then increases
Answer:
(b). Atomic size decreases along a period. However, the noble gas atom is an exception. It has very large size.

Question 26.
Which of the following set of elements is written in order of their increasing metallic character ?
(a) Be Mg Ca
(b) Na Li K
(c) Mg Al Si
(d) C O N
Answer:
(a). These elements belong to group 2. These are written in increasing order of their size. Since the metallic character increases down the group, the order is the correct.

NCERT Exemplar Solutions for Class 10 Science Chapter 5 Short Answer Questions

Question 27.
The three elements A, B and C with similar properties have atomic masses X, Y and Z respectively. The mass of Y is approximately equal to the average mass of X and Z. What is such an arrangement of elements called as ? Give one example of such a set of elements.
Answer:
The arrangement is known as Dobereiner’s triad. For example, Calcium (Ca), Strontium (Sr) and Barium (Ba).

Question 28.
Elements have been arranged in the following sequence on the basis of their increasing atomic masses.
F, Na, Mg, Al, Si, P, S, Cl, Ar, K
(a) Pick two sets of elements which have similar properties.
(b) The given sequence represents which law of classification of elements ?
Answer:
(a) The elements that have similar properties belong to the same group. From the list of elements available, elements which belong to same group are : .
Na, K (Alkali metals) ; F, Cl (Halogens)
(b) The sequence is according to Newland’s law of octaves.

Question 29.
Can the following groups of elements be classified as Dobereiner’s triad ?
(a) Na, Si, Cl
(b) Be, Mg, Ca
Answer:
Atomic mass of Be = 9; Na = 23; Mg = 24; Si = 28; Cl = 35; Ca = 40 Explain by giving reason.
(a) No, these elements cannot be classified as triads because these do not have same properties. However, the atomic mass of Si (28) is almost the mean of the atomic masses of elements Na (23) and Cl (35).
Note : The Dobereiner’s triad is meaningful only if the elements present in the triad have almost identical properties.
(b) Yes, these elements can be classified as triads. These belong to the same group (2) and have almost identical properties. The atomic mass of Mg (24) is almost the mean of the atomic masses of the elements Be (9) and Ca(40) i.e., 9 + 40 = 49/2 = 24.5.

Question 30.
In Mendeleev’s Periodic Table, the elements were arranged in the increasing order of their atomic masses. However, cobalt with atomic mass of 58.93 amu was placed before nickel having an atomic mass of 58.71 amu. Give reason for the same.
Answer:
This is regarded as a defect in the Mendeleev’s periodic table. However, the main reason for making the arrangement was that the elements with similar characteristics must be grouped together. In this case,
• Cobalt (Co) should be in the company of the elements Rhodium (Rh) and Iridium (Ir).
• Nickel (Ni) should be in the company of the elements Palladium (Pd) and Platinum (Pt).

Question 31.
“Hydrogen occupies a unique position in Modern Periodic Table”. Justify the statement.
Answer:
The position of the element hydrogen is still not clear even in the Modern Periodic Table.
• In electronic configuration, it resembles alkali metals of group 1. All of them have only one electron in the valence shell. Actually, hydrogen has only one shell (K-shell) which has one electron. However, it is not a metal whereas alkali-metals are metallic in nature.
• In characteristics, it resembles halogens of group 17. For example, like halogens hydrogen is a non-metal and diatomic as well.
It has been therefore, decided to assign hydrogen a unique position in the Modern or Long Form. Periodic Table. It is placed at the top in group 1 of alkali metals. However, it is not a member of that group.*

Question 32.
Write the formulae of chlorides of Eka-silicon and Eka-aluminium, the ^elements predicted by Mendeleev.
Answer:
Eka-aluminium represents gallium (Ga) with valency three and Eka-silicon is for germanium (Ge) with valency four. The formulae of their respective chlorides are GaCl₃ and GeCl₄.

Question 33.
Three elements A, B and C have 3, 4 and 2 electrons respectively in their outermost shell. Give the group number to which they belong in the Modern Periodic Table. Also, give their valencies.
Answer:
The electrons present in the outermost shell are also known as valence electrons. The desired information is given :

Question 34.
If an element X is placed in group 14, what will be the formula and the nature of bonding of its chloride ?
Answer:
The element X present in group 14 has four valence electrons in its atom. It can complete its octet by sharing four valence electrons with the electrons of other atoms. Therefore, it will form covalent bonds with the four atoms of chlorine. The formula of the chloride of the element X is.

Question 35.
Compare the radii of two species X and Y. Give reasons for your answer.
(a) X has 12 protons and 12 electrons
(b) Y has 12 protons and 10 electrons
Answer:
The available information makes it clear that :
(a) Species X with equal number of protons (12) and electrons (12) is an atom.
(b) Species Y with two electrons less (10) than the number of protons (12), is a divalent cation y²⁺ of species X.
Therefore, the radius of species X is more as compared to that of y²⁺.
Note : The radius of cation is always less than that of atom while that of anion is more.

Question 36.
Arrange the following elements in increasing order of their atomic radii.
Answer:
(a) All the elements belong to the same period (second). The atomic size or radius decreases along a period. Therefore, increasing order of atomic radii is : F < N < Be < Li
(b) The listed elements are present in group 17 (halogen family). The atomic size or radius increases down
a group. Therefore, the correct increasing order of atomic radii is : Cl < Br < I < At.

Question 37.
Identify and name the metals out of the following elements whose electronic configurations are given below :
(a) 2, 8, 2
(b) 2, 8, 1
(c) 2, 8, 7
(d) 2, 1.
Answer:
(a) Electronic configuration : 2, 8, 2 (Mg is a metal)
(b) Electronic configuration : 2, 8, 1 (Na is a metal)
(c) Electronic configuration : 2, 8, 7 (Cl is a non-metal)
(d) Electronic configuration : 2, 1 (Li is a metal)

Question 38.
Write the formula of the compound formed when the element A (atomic number 19) combines with the element B (atomic number 17). Draw its electronic dot structure. ^What is the nature of the bond formed ? (CBSE 2013)
Answer:
Electronic configuration of element A (Z = 19) is 2, 8, 8, 1
Electronic configuration of element B(Z = 17) is 2, 8, 7.
The element A has one valence electron while the element B has seven electrons in its valence shell. One electron gets transferred from the atom of element A to the atom of element B. As a result of electron transfer, an ionic bond is formed.

Question 39.
Arrange the following elements in the increasing order of their metallic character. ,
Mg, Ca, K, Ge, Ga
Answer:
In general, the metallic character increases down the group and decreases along a period. From the relative positions of the elements in the periodic table, the increasing order of metallic character is :
Ge < Ga < Mg < Ca < K.

Question 40.
Identify the elements with the following property and arrange them in increasing order of their reactivity
(a) An metal which is soft and reactive
(b) The metal which is an important constituent of limestone
(c) The metal which exists in liquid state at room temperature
Answer:
(a) The metal may be either Na or K.
(b) The metal is Ca.
(c) The metal is Hg.

Question 41.
The increasing order of reactivity of the metals is :
Hg < Ca < Na < K.
Properties of the elements are given below. Where would you locate the following elements in the periodic table ?
(a) A soft metal stored under kerosene
(b) An element with variable (more than one) valency stored under water
(c) An element which is tetravalent and forms the basis of organic chemistry
(d) An element which is an inert gas with atomic number 2
(e) An element whose thin oxide layer is used to make other elements corrosion resistant by the process of “anodising.”
Answer:
(a) Sodium (Group 1 and Period 3) or Potassium (Group 1 and Period 4)
(b) Phosphorus (Group 15 and Period 3). It shows variable valencies 3 and 5 and is stored under water.
(c) Carbon (Group 14 and Period 2)
(d) Helium (Group 18 and Period 1)
(e) Aluminium (Group 13 and Period 3)

NCERT Exemplar Solutions for Class 10 Science Chapter 5 Long Answer Questions

Question 42.
An element placed in 2nd group and 3rd Period of the Periodic Table, burns in the presence of oxygen to form a basic oxide.
(a) Identify the element
(b) Write the electronic configuration
(c) Write the balanced equation when it burns in the presence of air
(d) Write a balanced equation when this oxide is dissolved in water
(e) Draw the electron dot structure for the formation of this oxide
Answer:
(a) The element is magnesium (Mg).
(b) The electronic configuration is 2, 8, 2.
(c) Magnesium burns in oxygen (air) to form magnesium oxide which is of basic nature.
2Mg(s) + O₂(g) ————-> 2MgO(s)
(d) Magnesium hydroxide is formed.
MgO(s) + H₂O(aq) ————> Mg(OH)₂(s)

Question 43.
An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a divalent halide.
(a) Where in the Periodic Table are elements X and Y placed ?
(b) Classify X and Y as metal(s), non-metal(s) or metalloid(s).
(c) What will be the nature of oxide of element Y ? Identify the nature of bonding in the compound formed
(d) Draw the electron dot structure of the divalent halide.
Answer:
Element X (Z = 17) is chlorine while element Y (Z = 20) is calcium.
(a) Chlorine (Cl) is a member of group 17 and period 3. Calcium (Ca) is present in group 2 and period 4.
(b) Chlorine is a typical non-metal while calcium is a metal.
(c) The oxide of calcium is calcium oxide (CaO). It is a basic oxide.

Question 44.
Atomic number of a few elements are : 10, 20, 7, 14
(a) Identify the elements
(b) Identify the Group number of these elements in the Periodic Table
(c) Identify the Periods of these elements in the Periodic Table
(d) What would be the electronic configuration for each of these elements ?
(e) Determine the valency of these elements.
Answer:
(a) The elements are : Neon (Z = 10), Calcium (Z = 20), Nitrogen (Z = 7) and Silicon (Z = 14)
(b) Group numbers : Neon (18), Calcium (2), Nitrogen (15), Silicon (14).
(c) Periods : Neon (2), Calcium (4), Fluorine (2), Silicon (3).
(d) Electronic Configuration : Neon (2, 8) ; Calcium (2, 8, 8, 2) Nitrogen (2, 5) ; Silicon (2, 8, 4)
(e) Valency : Neon (zero) ; Calcium (2) ; Nitrogen (3) ; Silicon (4)

Question 45.
(a) In this ladder symbols of elements are jumbled up. Rearrange these symbols of elements in the increasing order of their atomic number in the Periodic Table.
(b) Arrange them in the order of their group also.

Answer:
(a) H, He, Li, Be, B, C, N, O, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca, Br.
(b) Group 1 — H, Li, Na, K
Group 2 — Be, Mg, Ca
Group 13 — B, A1
Group 14 — C, Si
Group 15 — N, P
Group 16 — O, S
Group 17 — Cl, Br
Group 18 — He, Ne, Ar

Question 46.
Complete the following cross word puzzle
Across :
(1) An element with atomic number 12.
(3) Metal used in making cans and member of Group 14.
(4) A lustrous non-metal which has 7 electrons in its outermost shell.
Down :
(2) Highly reactive and soft metal which imparts yellow colour when subjected to flame and is kept in kerosene.
(5) The first element of second period
(6) An element which is used in making fluorescent bulbs and is second member of group 18 in the Modern Periodic Table.
(7) A radioactive element which is the last member of the halogen family.
(8) Metal which is an important constituent of steel and forms rust when exposed to moist air.
(9) The first metalloid in Modern Periodic Table whose fibres are used in making bullet proof vests.

	1	7				2
		3	8				9		5
						4				6

                             (Cross-Puzzle)
Answer:

	1M	7A	G	N	E	2s	I	U	M
		S				O
		3T	8I	N		D	9B		5L
		A	R			4I	o	D	I	6n	E
		T	O			U	R		T	E
		I	N			M	O		H	O
		N					N		I	N
		E							U
									M

                               (Puzzle Solved)

Question 47.
Mendeleev predicted the existence of certain elements not known at that time and named two of them as Eka-silicon and Eka-aluminium.
(a) Name the elements which have taken the place of these elements.
(b) Mention the group and the period of these elements in the Modern Periodic Table.
(c) Classify these elements as metals, non-metals or metalloids.
(d) How many valence electrons are present in each one of them ?
Answer:
(a) Germanium (Ge) for Eka-silicon and gallium (Ga) for Eka-aluminium
(b) Germanium (Group 14 and Period 4)
Gallium (Group 13 and Period 4)
(c) Germanium (Metalloid) ; Gallium (Metal)
(d) Germanium (Z = 32), Four valence electrons (2, 8, 18, 4)
Gallium (Z = 31) ; Three valence electrons (2, 8, 18, 3).

Question 48.
(a) Electropositive nature of the element(s) increases down the group and decreases across the period
(b) Electronegativity of the element decreases down the group and increases across the period
(c) Atomic size increases down the group and decreases across a period (left to right)
(d) Metallic character increases down the group and decreases across a period.
On the basis of the above trends of the Periodic Table, answer the following about the elements with atomic numbers 3 to 9.
(a) Name the most electropositive element among them.
(b) Name the most electronegative element.
(c) Name the element with smallest atomic size.
(d) Name the element which is a metalloid.
(e) Name the element which shows maximum valency.
Answer:
(a) Lithium (Z = 3) is the most electropositive element,
(b) Fluorine (Z = 9) is the most electronegative element.
(c) Fluorine (Z = 9) has the smallest atomic size.
(d) Boron (Z = 5) is a metalloid.
(e) Carbon (Z = 6) shows the maximum valency (4). However, the element nitrogen (Z = 7) can show valency 5 in some compounds (e.g., N₂O₅).

Question 49.
An element X which is a yellow solid at room temperature shows catenation and allotropy. X forms two oxides which are also formed during the thermal decomposition of ferrous sulphate crystals and are the major air pollutants.
(a) Identify the element X.
(b) Write the electronic configuration of X.
(c) Write the balanced chemical equation for the thermal decomposition of ferrous sulphate crystals.
(d) What would be the nature (acidic/basic) of oxides formed ?
(e) Locate the position of the element in the Modern Periodic Table.
Answer:
(a) The available information suggests that the element X is sulphur.
(b) Electronic configuration of S(Z = 16) 2, 8, 6

(d) Fe₂O₃ (basic oxide), SO₂(acidic oxide), SO₃(acidic oxide)
(e) Sulphur is a member of group 16 and period 3.

Question 50.
An element X of group 15 exists as diatomic molecule and combines with hydrogen at 773 K in presence of the catalyst to form a compound, ammonia which has a characteristic pungent smell.
(a) Identify the element X. How many valence electrons does it have ?
(b) Draw the electron dot structure of the diatomic molecule of X. What type of bond is formed in it ?
(c) Draw the electron dot structure for ammonia. What type of bonds is formed in it ?
Answer:
(a) The available information suggests that the element X is nitrogen (N) and exists in diatomic form as N,
Electronic configuration of N(Z = 7) ; 2, 5. It has five valence electrons.

Question 51.
Which group of elements could be placed in Mendeleev’s Periodic Table without disturbing the original order ? Give reason.
Answer:
Group of noble gases (called zero group) could be placed in the Mendeleev’s Periodic Table with out disturbing the original order.
Reasons : Elements present in a group have same valency in the Mendeleev’s Periodic Table. Based on their electronic configuration, the members of noble gas family have zero valency. That is why they are called inert gases. They could be easily placed in the Mendeleev’s Periodic Table as a separate group without disturbing the arrangement of other elements.

Question 52.
Give an account of the process adopted by Mendeleev for the classification of elements. How did he arrive at “Periodic Law” ?
Answer:

1. The basis of classification of elements adopted was atomic masses of the elements. The elements were arranged in order of increasing atomic masses.
2. Elements with similar properties were kept in a particular group in order of increasing atomic masses.
3. Elements placed in a particular group (or sub group) were having same valency.

Note : Please note that when Mendeleev arranged the elements in the Periodic Table only 63 elements were known. Even noble gas or inert gas elements were adjusted at a later stage. Moreover, many groups were left in the table.

 

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HOTS Questions for Class 10 Science Chapter 5 Periodic Classification of Elements

Question 1.
Atoms of eight elements A, B, C, D, E, F, G and H have the same number of electronic shells but different number of electrons in their outermost shell. It was found that elements A and G combine to form an ionic compound. This compound is added in a small amount to almost all vegetable dishes during cooking Oxides of elements A and B are basic in nature while those of E and F are acidic. The oxide of D is almost neutral. Based on the above information answer the following questions :

1. To which group or period of the periodic table do the listed elements belong ?
2. What would be the nature of compound formed by a combination of elements B and F ?
3. Which two elements could definitely be metals ?
4. Which one of the eight elements is most likely to be found in gaseous state at room temperature ?
5. If the number of electrons in the outermost shell of elements C and G be 3 and 7 respectively, write the formula of the compound formed by the combination of C and G.

Answer:
The clue for the answer is given by the compound which is added to almost all vegetables during cooking. It is sodium chloride (NaCl). Both these elements belong to third period :

1. All the elements belong to third period. They have three shells (K, L, M) and the number of electrons vary from 1 to 8. These belong to different groups which are listed.
   
2. The element ‘B’ is Ca and ‘F’ is S. They combine to form CaS. It is an ionic compound.
3. The elements A, B are definite metals because the elements present in group 1 and 2 are all metals.
4. The element H is a noble gas element. It is most likely to be found in the gaseous state at room temperature.
5. The electronic configuration of the element C(Z = 13) is 2, 8, 3 while that of G(Z = 17) is 2, 8, 7. The formula of the compound formed by their combination is CG₃. It is actually AlCl₃ and is formed as a result of electron sharing.
   

More Resources

• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
The following table shows the position of six elements A, B, C, D, E and F in the period table.

Groups Periods	I	2	3 to 12	13	14	15	16	17	18
2.		A				B			C
3.		D			E				F

Using the above table answer the following questions :
(a) Which element will form only covalent compounds ?
(b) Which element is a metal with valency 2 ?
(c) Which element is a non-metal with valency of 3 ?
(d) Out of D and E, which one has more atomic radius and why ?
(e) Write a common name for the family of elements C and F.
Answer:
(a) The element ‘E’ present in group 14 is a non-metal. Its name is silicon (Si) and the compounds of the element are only covalent.
(b) The element ‘D’ present in group 2 is a metal known as magnesium (Mg). It exhibits valency 2 in its compounds.
(c) The elements ‘B’ present in group 15 is a non-metal. It is nitrogen (N) and exhibits valency 3 in its compounds.
(d) The element ‘D’ has more atomic radius than the element ‘E’ as the atomic size decreases along a period.
(e) The elements ‘C’ and ‘F’ present in group 18 belong to a family known as noble gases.

Question 3.
Two elements X and Y belong to group 1 and 2 respectively in the same period. Compare them with respect to :
(a) the number of valence electrons
(b) valency
(c) metallic character
(d) size of the atoms
(e) formulae of their oxides and chlorides.
Answer:
(a) The valence electrons present in element X (group 1) and element Y (group 2) are 1 and 2 respectively.
(b) The valency of the element X is one while that of the element Y is two.
(c) Metallic character decreases along a period. This means that the element X is more metallic as compared to element Y.
(d) Atomic size decreases along a period. As a result, the element Y has a smaller size than the element X.
(e) For element X : oxide (X₂O), chloride (XCl).
For element Y : oxide (YO) and chloride (YCl₂).

Question 4.
Atoms of seven elements A, B, C, D, E, F and G have a different number of electronic shells but have the same number of electrons in their outermost shells. The elements A and C combine with chlorine to form an acid and common salt respectively. The oxide of element A is liquid at room temperature and is a neutral substance while the oxides of the remaining six elements are basic in nature. Based on the above information, answer the following questions given ahead :

1. What could the element A be ?
2. Will elements A to G belong to the same period or same group of the periodic table ?
3. Write the formula of the compound formed by the reaction of the element A with oxygen.
4. Show the formation of the compound by a combination of element C with chlorine with the help of electronic structure.
5. What would be the ratio of number of combining atoms in a compound formed by the combination element A with carbon ?
6. Which one of the given elements is likely to have the smallest atomic radius ?

Answer:
Since all the seven elements have same number of electrons in their outer most shells, this means that they belong to the same group. From the available information, it becomes clear that the group is of alkali metals (group 1). Hydrogen (A) is also included. The elements present are;

Let us answer the questions asked

1. The element A’ is hydrogen (H)
2. The elements A’ to ‘G’ belong to the same group and not same period
3. The formula of compound is H₂O
4. Compound between ‘C’ and chlorine (Cl) is sodium chloride (NaCl) formed by the transference of one electron.
   
5. The compound formed between A (H) and carbon is methane (CH₄). In this the two elements are present in the ratio of 1 : 4
6. The element A (hydrogen) has the smallest atomic radius since it is the first element of the group.

 

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