Category: CBSE

NCERT Solutions for Class 12 Chemistry Chapter 2 contains solved questions asked in the textbook. The answers are provided by subject experts and hence can be used for reference. Chemistry is an important subject for boards as well as competitive exams. NCERT Solutions are therefore the best guide for the students appearing for such exams.

Every minute detail is explained in such a way that the students find it easy to understand. The students from different boards such as UP board, MP board, CBSE, Gujarat board can refer to the NCERT Solutions for Class 12 Chemistry Chapter 2 to score well in the exams.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 2
Chapter Name	Solutions
Number of Questions Solved	52
Category	NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions

Class 12 Chemistry Chapter 2 Solutions are defined as homogeneous mixture of two or more components. This chapter gives an overview of different types of solutions. Various laws and its derivations are provided here stepwise for easy understanding. The NCERT Solutions for Class 12 Chapter 2 provide solutions to the questions provided in the textbook.

NCERT IN-TEXT QUESTIONS

Question 1.
Calculate the mass percent of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Answer:

Question 2.
Calculate the mole fraction of benzene in a solution containing 30% by mass of it in carbon tetrachloride.
Answer:
Let us start with 100 g of the solution in which
Mass of benzene = 30 g
Mass of carbon tetrachloride = 70 g
Molar mass cf benzene (C₆H₆) = 6 x 12 + 6 x 1 = 78g mol^-1.

Question 3.
Calculate the molarity of each of the following solutions:
(a) 30 g of CO(NO₃)₂.6H₂O in 4.3 L of solution
(b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.
Answer:
(a) Molar mass of CO(NO₃)₂.6H₂O=310.7 g mol^-1
no. of moles = 30/310.7 = 0.0966
Vol. of solution = 4.3 L
Molarity =0.0966/4.3 = 0.022M
(b) 1000 mL of 0.5M  H₂SO₄ contain H₂SO₄ = 0.5 mole
30 mL of 0.5 M H₂SO₄ contain H₂SO₄
=0.5/1000 x 30 = 0.015 mole
Volume of solution = 500mL=0.5 L
Molarity = 0.015/0.5 = 0.03M

Question 4.
Calculate the mass of urea (NH₂CONH₂) required to prepare 2-5 kg of 0-25 molal aqueous solution.
Answer:

Question 5.
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL^-1.
Answer:

Question 6.
H₂S a toxic gas with rotten egg smell, is used for the qualitative analysis. If the solubility of H₂S in water at S.T.P is 0·195 m ; calculate Henry’s law constant.
Answer:
Step I. Calculation of mole fraction of H₂S

Step II. Calculation of Henry’s Law constant
According to Henry’s Law,

Question 7.
Henry’s Law constant for CO₂ in water is 1·67 x 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2·5 atm pressure of CO₂ at 298 K. (D.S.B. 2008 Supp.)
Answer:
Step I. Calculation of number of moles of CO₂.
According to Henry’s Law,


(\({ n }_{ { CO }_{ 2 } }\) has been neglected as the gas is very little soluble in water)
∴ \({ n }_{ { CO }_{ 2 } }\) = \({ x }_{ { CO }_{ 2 } }\) x (27·78 mol) = (1·52 x 10^-3) x (27·78 mol) = 0·0422 mol
Step II. Mass of CO₂ dissolved in water = (0·0422 mol) x (44 g mol^-1) = 1·857 g.

Question 8.
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Answer:

Question 9.
The vapour pressure of pure water at 298 K is 23·8 mm Hg. 50 g of Urea (NH₂CONH₂) is added to 850 g of water. Calculate the vapour pressure of water for this solution and also it’s relative lowering in vapour pressure.
Answer:
Step I. Calculation of vapour pressure of water for the solution
According to Raoult’s Law,

Step II. Calculation of relative lowering in vapour pressure

Question 10.
The boiling point of water at 750 mm Hg is 99·63°C. How much sucrose is to be added to 500 g of water so that it may boil at 100°C? (K6 for water = 0·52 K kg mol^-1).
Answer:

Question 11.
Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K_f= 3.9 K kg mol^-1.
Answer:

Question 12.
Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1·0 g of a polymer of molar mass 185,000 in 450 mL of solution at 37°C.
Answer:

NCERT EXERCISE

Question 1.
Define the term solution. What kinds of solutions are possible? Write briefly about each type of solution with an example.
Answer:
A true solution is a homogenous mixture of two or more substances. The constituent particle which is in larger amount’’ is called a solvent and that in smaller quantity is called a solute.
TYPES OF SOLUTION

Question 2.
Give an example of a solid solution in which the solute is a gas.
Answer:
Solid in solid type. E.g: Copper in gold. This type of solutions are called alloys.

Question 3.
Define the following terms :
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Answer:
(i) Mole fraction :
The ratio of the number of moles of one component to the total number of moles of all the components present in the solution.
For a binary solution made up of components A and B,

(ii) Molality:
The number of gram moles of the solute dissolved in 1000 g (or kg) of the solvent.

(iii) Molarity:
The number of gram formula mass of the solute dissolved per litre of the solution.

(iv) Mass percentage:
The number of parts by mass of one component (solute or solvent) per 100 parts by mass of the solution. If A and B are the two components of a binary solution,

Question 4.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL^-1?
Answer:
68% nitric acid by mass means that 68g mass of nitric acid is dissolved in 100g mass of solution. Molar mass of HNO₃= 63g mol^-1

Question 5.
A solution of glucose in water is labelled as 10 percent W/W. What would be the molality and mole fraction of each component in the solution? If the density of the solution is 1·2 g mL^-1, then what should be the molarity of the solution? (C.B.S.E. 2013, Manipur Board 2015)
Answer:

Question 6.
How many mL of a 0·1 M HCl are required to react completely with a 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of two?
Answer:

Question 7.
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Answer:

Question 8.
An antifreeze solution is prepared from 222·6 g of ethylene glycol C₂H₄(OH)₂ and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1·072 g mL^-1, then what shall be the molarity of the solution? (C.B.S.E. Delhi 2007)
Answer:

Question 9.
A sample of drinking water was found to be severely contaminated with chloroform(CHCl₃) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in a water sample.
Answer:

Question 10.
What role does the molecular interaction play in a solution of alcohol and water?
Answer:
Alcohols dissolve in water due to the formation of intermolecular H-bonding with water.

Question 11.
Why do gases nearly always tend to be less soluble in liquids as the temperature is raised?
Answer:
The dissolution of a gas in a liquid is exothermic in nature because the gas contracts in volume.

Gas + Liquid ⇌ Dissolved gas ; ∆H = – ve

An increase in temperature will favour the reverse process since it is of endothermic nature. Therefore, the solubility of the gas in the solution decreases with the rise in temperature.

Question 12.
State Henry’s law and mention some important applications.
Answer:
Henry’s law: According to this law, ‘The mass of a gas dissolved per unit volume of a solvent at a constant temperature, is proportional to the pressure of the gas with which the solvent is in equilibrium’.

Let in unit volume of solvent, the mass of the gas dissolved is m and equilibrium pressure is P, then m α P or m = KP, where K is a constant. We can understand Henry’s law by taking the example of soda water bottle. Soda water contains carbon dioxide dissolved in water under pressure.

Applications of Henry’s law:

1. In the production of carbonated beverages: To increase the solubility of CO₂ in soft drinks, soda water, bear etc. the bottles are sealed at high pressure.

2. In exchange of gases in the blood: The partial pressure of O₂ is high in inhaled air, in lungs it combines with hemoglobin to form oxyhemoglobin. In tissues, the partial pressure of oxygen is comparatively low therefore oxyhemoglobin releases oxygen in order to carry out cellular activities.

3. In deep-sea diving: Deep-sea divers depend upon compressed air for breathing at high pressure underwater. The compressed air contains N₂ in addition to O₂, which are not very soluble in blood at normal pressure. However, at great depths when the diver breathes in compressed air from the supply tank, more N₂ dissolved in the blood and in other body fluids because the pressure at that depth is far greater than the surface atmospheric pressure. When the divers come towards the surface at atmospheric pressure, this dissolves nitrogen bubbles out of the blood. These bubbles restrict blood flow, affect the transmission of nerve impulses. This causes a disease called bends or decompression sickness. To avoid bends, as well as toxic effects of high concentration of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% He, 56.2% N_2, and 32.1% O₂).

4. At high altitudes: At high altitudes, the partial pressure of O₂ is less than that at the ground level. This results in a low concentration of oxygen in the blood and tissues of the people living at high altitudes or climbers. The low blood oxygen causes climbers to become weak and unable to think clearly known as anoxia.

5. Aquatic life: The dissolution of oxygen (from air) in water helps in the existence of aquatic life in various water bodies like Lake, rivers, and sea.

Question 13.
The partial pressure over a saturated solution containing 6·56 x 10^-2 g of ethane is 1 bar. If the solution contains 5·0 x 10^-2 g of ethane, what shall be the partial pressure of the gas?
Answer:
According to Henry’s law,
The mass of the gas (m) dissolved in solution ∝ Partial pressure (p) (At constant temperature)

Question 14.
What is meant by positive and negative deviations from Raoult’s law and how is the sign of ∆H_sol related to positive and negative deviations from Raoult’s law?
Answer:
(a) Positive Deviation:
(i) ∆V_mixing is positive: This is quite likely also because in ge presence of weak forces of interaction, interaction, the volume of the solution is bound to increase.
(ii) ∆H_mixing is positive: Energy is needed to form the solution because the components of the solution have to be brought closer to form the solution. Thus, the process of mixing is of endothermic nature.

(b) Negative Deviation:
(i) ∆V_mixing is negative: Because of the increased forces of interaction, the molecules of the two components will come closer and as a result, there is a decrease in the volume of the solution.
(ii) ∆H_mixing is negative: Energy is expected to be released because of the increase in the forces of interaction. Therefore, the process of mixing is exothermic in nature or ∆H_mixing is negative.

Question 15.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Answer:

Question 16.
Heptane and octane form ideal solutions. At 373 K, the vapour pressure of the two liquid components is 105·2 k Pa and 46·8 k Pa respectively. If the solution contains 25 g of heptane and 35 g of octane, calculate:
(i) Vapour pressure exerted by heptane
(ii) Vapour pressure exerted by octane
(iii) Vapour pressure exerted by the solution
(iv) Mole fraction of octane in the vapour phase. (C.B.S.E. Sample Paper, 2010)
Answer:

Question 17.
The vapour pressure of water is 12·3 kPa at 300 K. Calculate the vapour pressure of 1 molal solution in it.
Answer:
1 molal solution implies one mole of the solute dissolved in 1000 g (1 kg) of solvent i.e. water.

Question 18.
Calculate the mass of a non-volatile solute (molar mass 40 g mol^-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Answer:

Question 19.
A solution containing 30 g of a non-volatile solute exactly in 90 g of water has a vapour pressure of 2·8 k Pa at 298 K. Further 18 g of water is then added to the solution and the new vapour pressure becomes 2·9 k Pa at 298 K. Calculate
(i) Molecular mass of the solute.
(ii) Vapour pressure of water at 298 K. (C.B.S.E. Outside Delhi 2005)
Answer:

Question 20.
A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of a 5% glucose in water if the freezing point of pure water is 273·15 K. (C.B.S.E. Delhi 2008)
Answer:


Freezing point temperature of glucose solution = (273·15 – 4·085) K = 269·07 K.

Question 21.
Two elements A and B form compounds having molecular formulae AB₂ and AB₄. When dissolved in 20 g of benzene, 1 g of AB₂ lowers the freezing point by 2·3 K whereas 1 g of AB₄ lowers it by 1·3 K. Molal depression constant for benzene is 5·1 K kg mol^-1. Calculate atomic masses of A and B. (C.B.S.E. Delhi 2004)
Answer:

Question 22.
At 300 K, 36g of glucose present in a litre of its solution has an osmotic pressure of 4.08 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Answer:

Question 23.
Suggest the most important type of intermolecular attractive interactions in the following pairs :

1. n-hexane and n-octane
2. I₂ and CCl₄
3. NaClO₄ and water (H₂O)
4. methanol and acetone
5. acetonitrile (CH₃CN) and acetone (C₃H₆O).

Answer:

1. Both are non-polar. Hence, intermolecular interactions in them will be London/ dispersion forces (discussed in class XI)
2. Both are non-polar. Hence, intermolecular interactions in them will be London/ dispersion forces (discussed in class XI)
3. NaClO₄ gives Na+ and ClO₄^– ions in the solution while water is a polar molecule. Hence, intermolecular interactions in them will be ion-dipole interactions.
4. Both are polar molecules. Hence intermolecular interactions in them will be dipole-dipole interactions.
5. Both are polar molecules. Hence intermolecular interactions in them will be dipole-dipole interactions.

Question 24.
Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain.
Cyclohexane, KCl, CH₃OH, CH₃CN.
Answer:
(i) Cyclohexane and n-octane both are non-polar. Hence they mix completely in all proportions.
(ii) KCl is an ionic compound while n-octane is nonpolar. Hence, KCl will not dissolve at all in n-octane.
(iii) CH₃OH and CH₃CN both are polar but CH₃CN is less polar than CH₃OH. As the solvent is non-polar, CH₃CN will dissolve more than CH₃OH is n-octane.
Thus the order of solubility will be KCl< CH₃OH < CH₃CN < Cyclohexane.

Question 25.
Among the following compounds, identify which are insoluble, partially soluble, and highly soluble in water?
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol.
Answer:
(i) phenol (C₆H₅OH): Is partially soluble in water due to weak dipole-dipole interactions in the molecules of phenol and water.
(ii) toluene (C₇H₈): Is insoluble in water because it is an aromatic hydrocarbon (non-polar) while water is polar in nature.
(iii) formic acid (HCOOH): Is highly soluble in water since it can form hydrogen bonding with water.
(iv) ethylene glycol (HOCH₂CH₂OH): Is highly soluble in water since it can form hydrogen bonding with water.
(v) chloroform (CHCl₃): Is insoluble in water because it is an organic heavy liquid and forms a separate layer.
(vi) pentanol (C₅H₁₁OH): In partially soluble in water because the bulky C5H11 group decreases its extent of hydrogen bonding with water.

Question 26.
If the density of some lake water is 1.25 g mL^-1 and contains 92g of Na⁺ ions per kg of water, calculate the molality of Na⁺ ions in the lake.
Answer:

Question 27.
If the solubility product of CuS is 6 x 10^-16, calculate the maximum molarity of CuS in aqueous solution.
Answer:
Dissociation of CuS in aqueous solution is :

By definition, Ksp corresponds to the product of the ionic concentration of the salt in saturated solution and it represents the maximum molarity of the salt. Therefore, maximum molarity of the salt = 2· 45 x 10^-8 M.

Question 28.
Calculate the mass percent of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6·5 g of aspirin is dissolved in 450 g of CH₃CN.
Answer:

Question 29.
Nalorphene (C₁₉H₂₁NO₃) similar to morphine is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1· 5 mg. Calculate the mass of 1· 5 x 10^-3 m aqueous solution required for the above doze?
Answer:

Question 30.
Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol 0.15 M solution means that 0.15 mole of benzoic acid is dissolved in 1L of solution.
Answer:

Question 31.
The depression in freezing point of water observed for the same amount of acetic acid, trichloroactetic acid and trifluoroacetic acid increases in the order given above. Explain. (C.B.S.E. 2008 Supp.)
Answer:
The depression in freezing point of a solute in water depends upon the number of particles or ions furnished by it in solution or upon its degree of dissociation (α). All the three organic acids ionise in aqueous solution. However, the relative order of acidic strengths is as given below.

This is linked with the electronegativity of the halogen atoms present. Fluorine (F) is more electronegative than (Cl). Under the circumstances, trifluoroacetic acid gives maximum ions in solution since it is the strongest acid. Consequently, the depression in freezing point (∆T_f) is the maximum in this case and is the least for acetic acid which is the weakest acid.

Question 32.
Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CH(Cl)COOH is added to 250 g of water. K_a = 1·4 x 10^-3; K_f = 1·86 K kg mol^-1. (C.B.S.E. 2008 Supp.)
Answer:
Step I. Calculation of degree of dissociation of acid Mass of acid = 10 g

Question 33.
19·5 g of CH₃FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1·0°C. Calculate Van’t Hoff factor and dissociation constant of the acid: K_f = 1·86 K kg mol^-1.
Answer:

Question 36.
100 g of liquid A (molar mass 140 g mol^-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^-1). The vapour pressure of pure liquid B was found to be 500 torrs. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.
Answer:

Question 37.
Vapour pressures of pure acetone and chloroform at 328 K are 632·8 mm Hg and 741·8 mm Hg respectively. Assuming that they form an ideal solution over die entire range of composition, plot p_total. p_chloroform, and p_acetone as a function of x_acetone. The experimental data observed for different compositions of the mixture is:

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative the ideal solution.
Answer:
From the available information:

Since the plot or graph dips downwards, the solution shows a negative deviation from Raoult’s Law.

Question 38.
Benzene and naphthalene form ideal solutions over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50·71 mm Hg and 32·06 mm Hg respectively. Calculate the mole pure fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of naphthalene.
Answer:

Question 39.
Air is a mixture of a number of gases. The major components are oxygen and nitrogen with the approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if Henry’s law constants for oxygen and nitrogen at 298 K are 3·30 x 10⁷ mm and 6·51 x 10⁷ mm respectively, calculate the composition of these gases in water.
Answer:

Question 40.
Determine the amount of CaCl₂ (i = 2·47) dissolved in 2·5 litre of water so that its osmotic pressure is 0·75 atm at 27°C.
Answer:
According to Van’t Hoff equation :

Question 41.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25°C, assuming that it is completely dissociated.
Answer:

We hope the NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Chemistry Chapter 5 are the practice guide for all the students. It contains solved questions provided by the subject matter experts. The solutions are accurate and the answers can be written in the exams.

NCERT Solutions are provided for reference in various boards (CBSE, UP board, Gujarat board, MP board). These help the students practice well before the examination and know their shortcomings. NCERT Solutions not only help the students score well in the board exams but also help them get through competitive exams.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 5
Chapter Name	Surface Chemistry
Number of Questions Solved	35
Category	NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 5 Surface Chemistry

Class 12 Chemistry Chapter 5 Surface Chemistry is an important chapter and explains the types of chemical reactions, its pocesses and mechanisms that take place at the surface of the material. The chapter also contains the details on various important concepts such as Tyndall effect, Brownian motion, and catalyst reactions. Notes on various topics like colloids, emulsions, electrodialysis, enzyme reactions, electrophoresis, etc. are also provided here for reference.

NCERT IN-TEXT QUESTIONS

Question 1.
Why are substances like platinum and palladium often used for carrying out the electrolysis of aqueous solutions ?
Answer:
The metals like platinum and palladium are used as inert electrodes for carrying out the process of electrolysis because these are not attacked by the ions involved in the process.

Question 2.
Why does physisorption decrease with increase in temperature ?
Answer:
Physisorption or physical adsorption of a gas on the surface of a solid is exothermic in nature.

When temperature is increased, the equilibrium gets shifted in the backward direction to neutralise the effect of increase in temperature. Consequently, physisorption decreases with the increase in temperature.

Question 3.
Why are powdered substances more effective . adsorbents than their crystalline forms?
Answer:
Powdered substances have greater surface area as compared to their crystalline forms. Greater the surface area, greater is the adsorption.

Question 4.
Why is it necessary to remove CO when ammonia is obtained by Haber’s process ?
Answer:
Carbon monoxide (CO) acts as a poison for the catalyst iron as well as promoter molybdenum which are used in the Haber’s process. Moreover, it is likely to combine with iron to form iron carbonyl Fe(CO)₅. Therefore, it is necessary to remove it from the reaction mixture by suitable means.

Question 5.
Why is ester hydrolysis slow in the beginning and becomes fast after sometime ?
Answer:
In ester hydrolysis, an acid and alcohol are formed as the products. For example,

Acid will release H⁺ ions in solution which act as catalyst (auto-catalysis) for the reaction. That is why, the hydrolysis is slow in the beginning and becomes faster later on.

Question 6.
What is the role of desorption in the process of catalysis?
Answer:
Desorption makes the surface of the solid- catalyst-free for fresh adsorption of the reactants on the surface.

Question 7.
What modification can you suggest for Hardy-Schulze Law?
Answer:
According to Hardy-Schulze Law, the ions carrying charge opposite to the charge on sol particles neutralise their charge and thus cause their coagulation or precipitation. The law takes into account the charge carried by the ion and not its size. Smaller the size of the ion more will be its polarising power. Thus, the law should be modified in terms of the polarising power of the flocculating ion or the ion causing the precipitation. The modified form of the law states that “Greater the polarising power of the flocculating ion added, greater is its power to cause precipitation.”

Question 8.
Why is it essential to wash the precipitate with water before estimating it quantitatively?
Answer:
Some amounts of the electrolyte are mixed to form the ppt. Some of these electrolytes remain adsorbed on the surface of the particles of the ppt. Hence, it is essential to wash the ppt with water to remove the sticking electrolytes (or any other impurities) before estimating it quantitatively.

NCERT EXERCISE

Question 1.
Distinguish between the meaning of terms adsorption and absorption. Give one example in each case.
Answer:
Differences between Adsorption and Absorption:

Adsorption:

1. It is a process as a result of which one substance gets concentrated only on the surface of the other.
2. The concentration of adsorbate on the surface of the adsorbent is different than in the bulk.
3. It is a surface phenomenon.
4. Example: Adsorption of water vapour on silica gel.

Absorption:

1. It is a process as a result of which one substance gets uniformly distributed in the volume of the other.
2. Concentration is uniform in the entire solid system.
3. It is a bulk phenomenon.
4. Example: Adsorption of water vapour by dry calcium chloride.

Question 2.
What is the difference between physisorption and chemisorption?
Answer:

Question 3.
Why is a finely divided substance more effective as an adsorbent ?
Answer:
With the increase in surface area of adsorbent adsorption increases. Thus, in the powdered state (finely divided substance) or in porous state surface area of metals is more. Therefore, adsorption is more in these states.

Question 4.
What are the factors which influence the adsorption of a gas on a solid ?
Answer:
There are various factors that affect the rate of adsorption of a gas on a solid surface.

• Nature of the gas: Easily liquefiable gases such as NH₃, HCl, etc. are adsorbed to a great extent in comparison to gases such as H₂, O_2, etc. This is because Van der Waal’s forces are stronger in easily liquefiable gases.
• The surface area of the solid: The greater the surface area of the adsorbent, the greater is the adsorption of a gas on the solid surface.
• Effect of pressure: Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore adsorption increases with an increase in pressure.
• Effect of temperature: Adsorption is an exothermic process. Thus in accordance with Leehatelie’s principle, the magnitude of adsorption decreases with an increase in temperature.

Question 5.
What is adsorption isotherm ? Distinguish between Freundlich adsorption isotherm and Langmuir adsorption isotherm.
Answer:
Adsorption isotherm represents the variation of the amount of the gas adsorbed and the corresponding pressure at a certain temperature. The mathematical forms of the two adsorption isotherms are :

The main points of distinction in the two adsorption isotherms are:

• Freundlich Adsorption isotherm is applicable to all types of adsorption whereas Langmuir Adsorption isotherm is applicable mainly to chemical adsorption or chemisorption.
• Freundlich adsorption isotherm fails at high pressure of the gas whereas Langmuir Adsorption isotherm can be applied under all pressures.

Question 6.
What do you understand by activation of adsorbent? How is it achieved?
Answer:
By activating an absorbent, we tend to increase the adsorbing power of the adsorbent. Some ways to activate an adsorbent are:

1. By increasing the surface area of the adsorbent. This can be done by breaking it into smaller pieces or powdering it.
2. Some specific treatments can also lead to the activation of the adsorbent.

For example, wood charcoal is activated by heating it between 650K and 1330K in vacuum pr air. It expels all the gases absorbed or adsorbed and thus, creates a space for the adsorption of gases.

Question 7.
What role does adsorption play in heterogeneous catalysis?
Answer:
Heterogeneous catalysis is generally carried on the surface of the finely divided metals of the transition series. Due to the availability of large surface area, the reacting species get adsorbed on the surface either by physical adsorption or by chemisorption. The adsorbed species get opportunity to mutually combine to form the products which are released or desorbed from the surface so as to accommodate more reacting species.

• Diffusion of the reactants on the surface of the catalyst.
• Some association between the catalyst surface and the reactants i. e., adsorption.
• The occurrence of the chemical reactions on the catalyst surface.
• Dissociation of the reaction products from the catalyst surface i.e., desorption.
• Diffusion of the products from the catalyst surface.

Question 8.
Give two chemical methods for the preparation of colloids.
Answer:
These are formed in two ways:

• Condensation methods
• Dispersion methods.

Condensation methods: The particles of the dispersed phase are very small in size. They have to be condensed suitably to be of colloidal size.
A colloidal solution of sulphur is obtained when H₂S gas is bubbled through the solution of oxidising agent like bromine water, sulphur dioxide, dilute HNO₃ etc.
Dispersion methods: In these methods, bigger particles of a substance (suspension) are broken into smaller particles of colloidal dimensions. The substance whose colloidal solution is to be prepared, is first ground to coarse particles. It is then mixed with the dispersion medium to get a suspension

Question 9.
How are colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?
Answer:
There are in all eight types of colloidal solutions.

Question 10.
Discuss the effect of pressure and temperature on the adsorption of gases on solids.
Answer:
Effect of pressure:
Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore adsorption increases with an increase in pressure.

Effect of temperature:
Adsorption is an exothermic process. Thus, in accordance with Le-chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

Question 11.
What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
Answer:
Lyophillic colloids (solvent loving) are those substances that directly pass into the colloidal state when brought- in contact with the solvent, e.g., proteins, starch, rubber, etc.
These sols are quite stable because of the strong attractive forces between the particles of disperse phase and the dispersion medium.
Lyophobic colloids (solvent hating) are those substances that do not form the colloidal sol readily when mixed with the dispersion medium.These sols are less stable than the lyophilic sols. Examples of lyophobic sols include sols of metals and their insoluble compounds like sulphides and hydroxides.
The stability of hydrophobic sol is only due to the presence of charge on the colloidal parties. If charge is removed, e.g., by addition of suitable electrolytes, the particles will come nearer to each other to form aggregate, i.e., they will coagulate and settle down. On the other hand, the stability of hydrophilic sol is due to charge as well as solvation of the colloidal particles. Thuf, for coagulation to occur easily both the mentioned factors have to be removed.

Question 12.
What is the difference between muitimolecular and macromolecular colloids ? Give one example of each. How are associated colloids different from these two types of colloids ? (C.B.S.E. 2008, 2009, 2010)
Answer:
Difference between multimolecular and macromolecular colloids
The main points of distinction are listed.

Muitimolecular colloids	Macromolecular colloids
1.The particle size is less than that of the colloidal range (< 103 pm) 2. They exist as aggregates of smaller particles. 3. These are mostly lyophobic colloids.	1. The particle size falls in the colloidal range (103 to 106 pm). 2. These are already macro molecular in nature. 3. These are mostly lyophilic colloids.

Colloidal sol of sulphur (Sg) is an example of multimolecular colloid while colloidal sol of starch represents macromolecular colloid.

Associated colloids also called micelles, are generally electrolytes. They exist as ions at low concentrations. However, above a particular concentration called critical micellear concentration (CMC) and above a particular temperature called Kraft temperature (T_k), these get associated and exhibit colloidal behaviour. Soap is a common example of associated colloids.

Multimolecular colloids: In these colloids, the individual particles consist of an aggregate of atoms or small molecules with molecular size less than 10³ pm. For example, gold sol consists of particles of various sizes having several atoms. Similarly, a sulphur sol consists of particles each having eight sulphur atoms (S_g). In these colloids, the particles are held by van der Waals’ forces.

Macromolecular colloids: In this type, the particles of the dispersed phase are sufficiently big in size (macro) to be of colloidal dimensions. These are normally polymers. A few naturally occurring macromolecules are starch, cellulose and proteins. The examples of artificial macromolecules are those of polythene, nylon, polystyrene, plastics etc.

Question 13.
What are enzymes? Write in brief the mechanism of enzyme catalysis.
Answer:
Enzymes are complex nitrogenous compounds which are produced by living plants and animals. In fact, these are proteins produced by living systems and catalyse certain biological reactions. These are, therefore, often known as bio-chemical catalysts and this phenomenon is known as bio-chemical catalysis.

The rate of enzyme-catalyzed reaction which is initially of first-order changes to zero-order as the concentration of substrate species on the catalyst surface increases.
Two models have been proposed by bio-chemists to explain the mechanism of enzyme catalyzed reactions. These are briefly discussed.

Question 14.
How are colloids classified on the basis of :
(a) physical states of components
(b) nature of dispersion medium
(c) interaction between the dispersed phase and dispersion medium?
Answer:
(a) Based on physical states of components. Based on the physical states of components i.e., dispersed phase and dispersion medium, there are eight types of colloidal solutions.

(b) Nature of dispersion medium. The dispersion medium can be either gas, liquid or solid. Based upon its nature, the colloids or colloidal solutions are of three types.

• Aerosols: Air or gases act as the dispersion medium
• Liquid sols: Liquids like water, alcohol or benzene act as the dispersion medium.
• Solid sols: Solid acts as the dispersion medium.

(c) Interaction between the dispersed phase and dispersion medium. Colloidal solutions are classified into two types. These are lyophilic and lyophobic sols.
(i) Lyophilic colloids: The colloidal solution in which the particles of the dispersed phase have a great affinity (or love) for the dispersion medium, are called lyophilic colloids. Such solutions are easily formed the moment the dispersed phase and the dispersion medium come in direct contact. e.g., sols of gum, gelatin, starch, etc.

(ii) Lyophobic colloids: The colloidal solutions in which the particles of the dispersed phase have no affinity or love, rather have hatred for the dispersion medium, are called lyophobic colloids. The solutions of metals like Ag and Au, hydroxides like Al(OH)₃ and Fe(OH)₃ and metal sulphides like As₂S₃ are examples of lyophobic colloids.

Question 15.
Explain what is observed
(i) when a beam of light is passed through a colloidal sol.
(ii) an electrolyte, NaCI is added to hydrated ferric oxide sol.
(iii) electric current is passed through a colloidal sol.
Answer:
(i) Scattering of light by colloidal particles takes place and the path of the light becomes visible (Tyndall effect).
(ii) The positively charged colloidal particles of Fe(OH)₃ get coagulated by the oppositely charged Cl^– ions provided by NaCl.
(iii) On passing electric current, the colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated. This is the electrophoresis process.

Question 16.
What are emulsions? What are their different types? Give an example of each type.
Answer:
(a) Oil-in-water emulsion (O/W type). In this case, the dispersed phase is oil while the dispersion medium is water. Milk is a common example in which liquid fats are dispersed in water. Similarly, if a few drops of nitrobenzene (oil) is added to water, an emulsion results. Vanishing cream is another example of this type.

(b) Water-in-oil emulsion (W/O type). In this type of emulsions, the dispersed phase is water while the dispersion medium is oil. Butter is an example of water in oil emulsion in which water is dispersed in oil. Cod liver oil and cold cream are the other examples of these emulsions.

Question 17.
What is demulsification? Name two demulsifiers.
Answer:
The process of separation of constituent liquids of an emulsion is called demulsification. Demulsification can be done by centrifuging or boiling.

Question 18.
The action of soap is due to demulsification and micelle formation. Comment.
Answer:
Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOO^– Na⁺ (e.g., sodium stearate CH₃(CH₂)]₁₆ COO^– Na⁺, which is a major component of many bar soaps). When dissolved in water, it dissociates into RCOO^– and Na+ ions. The RCOO^– ions, however, consist of two parts – a long hydrocarbon chain R (also called non-polar tail’) which is hydrophobic (water-repelling), and a polar group COO^– (also called polar ionichead’), which is hydrophilic (water loving).

The RCOO^– ions are, therefore, present on the surface with their COO^– groups in water and the hydrocarbon chains R staying away from it and remain at the surface. But at critical micelle concentration, the anions are pulled into the bulk of the solution and aggregate to form a spherical shape with their hydrocarbon chains pointing towards the centre of the sphere with COO^– part remaining outward on the surface of the sphere. An aggregate thus formed is known as ‘ionic micelle’.

The cleansing action of soap is due to the fact that soap molecules form micelle around the oil droplet in such a way that the hydrophobic part of the stearate ions is in the oil droplet and the hydrophilic part projects out of the grease droplet like the bristles

Question 19.
Give four examples of heterogeneous catalysts.
Answer:
(i) The combination between nitrogen and hydrogen to form ammonia in the presence of finely divided iron acting as catalyst. This is known as Haber’s process.

(ii) Formation of sulphur trioxide by the oxidation of sulphur dioxide in the presence of platinum catalyst is the basis of the manufacture of sulphuric acid in Contact process.

(iii) Oxidation of ammonia into nitric oxide in the presence of platinum catalyst is employed for the commercial preparation of nitric acid in Ostwald process.

(iv) In the hydrogenation of vegetable oils (unsaturated in nature) resulting in solid fats (saturated in nature), hydrogen gas is passed through the oil in the presence of nickel catalyst at about 473 K.

Question 20.
What do you mean by activity and selectivity of catalysts?
Answer:
(a) Activity: The activity of a catalyst depends upon the strength of chemisorption to a large extent. The reactants must get adsorbed reasonably strongly onto the catalyst to become active. But adsorption must not be so strong that they are immobilised. It is observed that maximum activity is shown by elements of groups 7 – 9 of the periodic table
2H₂ + O₂ \(\underrightarrow { Pt }\) 2H₂O

(b) Selectivity: The selectivity of a catalyst is its ability to yield a particular product in the reaction e.g.,

Thus, a selective catalyst can act as a catalyst in one reaction and may fail to catalyze another reaction.

Question 21.
Describe some features of catalysis by zeolites.
Answer:

1. Zeolites are hydrated alumino-silicates. They have a three-dimensional network structure. They contain water molecules in their pores,
2. Zeolites are heated to remove the water from hydration. The pores become vacant and zeolites are ready to act as catalysts.
3. The size of the pores varies from 260 pm to 760 pm. This shows that only those molecules can be adsorbed in these pores whose size is small enough to enter these pores. Thus, zeolites a molecular sieve and the shape-selective catalysts.

Question 22.
What are shape-selective catalysts?
Answer:
The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape-selective catalysis. Zeolites are good shape-selective catalysts because of their honeycomb-like structures. They are microporous aluminosilicates with a three-dimensional network of silicates in which some silicon atoms are replaced by aluminium atoms giving an Al-O-Si framework. The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules as well as upon the pores and cavities of the zeolites. They are found in nature as well as synthesized for catalytic selectivity.

Question 23.
Explain the following terms:
(i) Electrophoresis
(ii) Coagulation
(iii) Dialysis
(iv) Tyndall effect.
Answer:
(i) Electrophoresis:
The movement of colloidal particles under the influence of an electric field is known as electrophoresis. Positively charged particles move to the cathode, while negatively charged particles move towards the anode. As the particles reach oppositely charged electrodes, they become neutral and get coagulated.

(ii) Coagulation:
The process of setting down colloidal particles i.e., conversion of a colloid into a precipitate is called coagulation.

(iii) Dialysis: The process of removing dissolved substances from a colloidal solution by the means of diffusion through a membrane is known as dialysis. This process is based on the principle that ions and small molecules can pass through animal membranes, unlike colloidal particles.

(iv) Tyndall effect:
When a beam of light is allowed to pass through a colloidal solution, it becomes visible like a column of light. This is known as the Tyndall effect. This phenomenon takes place as particles of colloidal dimension scatter light in all directions.

Question 24.
Give four uses of emulsions.
Answer:
Four uses of emulsions:

1. the cleansing action of soaps is based on the formation of emulsions
2. digestion of fats in the intestines takes place by the process of emulsification.
3. Antiseptics and disinfectants when added to water form emulsions.
4. The process of emulsification is used to make medicines.

Question 25.
What are micelles? Give an example of the micelles system.
Answer:
Micelles are substances that behave as normal strong electrolytes at low concentration but at high concentrations behave as colloids due to the formation of aggregates. They are also called associated colloids, e.g., soaps and detergents. They can form ions and may contain 100 or more molecules to form a micelle.

Question 26.
Explain the terms with suitable examples:

1. Alcosol
2. Aerosol
3. Hydrosol.

Answer:

1. Alcosol: It is a colloidal solution in which alcohol is the dispersion medium. For example, colloids which has cellulose nitrate as a dispersed phase and ethyl alcohol as the dispersion medium.
2. Aerosol: It is a colloidal solution in which liquid is a dispersed phase and gas is a dispersion medium e.g., fog, mist, cloud, etc.
3. Hydrosol: It is a colloidal solution in which solid is a dispersed phase and water is a dispersion, e.g., gold sol, arsenious sulphide sol, ferric oxide sol, etc.

Question 27.
Comment on the statement that colloid is not a substance but a state of substance’.
Answer:
Common salt (atypical crystalloid in an aqueous medium) behaves as a colloid in a benzene medium. Hence, we can say that a colloidal substance does not represent a separate class of substances. When the size of the solute particle lies between 1 nm and 1000 nm, it behaves as a colloid.

Hence, we can say that colloid is not a substance but a state of the substance which is dependent on the size of the particle.
A colloidal state is intermediate between a true solution and a suspension.

We hope the NCERT Solutions for Class 12 Chemistry Chapter 5 Surface Chemistry help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 5 Surface Chemistry, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Chemistry Chapter 6 contains concept wise details of the chapter. The solutions help the students prepare well for the examination. The diagrammatic representations and step wise solutions make it easy for the students to understand. Also the answers are accurate and are provided by the subject matter experts.

NCERT Solutions are provided for reference in various boards (CBSE, MP board, UP board, Gujarat board). The students appearing for these boards and competitive exams can score well by referring to the NCERT Solutions.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 6
Chapter Name	General Principles and Processes of Isolation of Elements
Number of Questions Solved	31
Category	NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

Class 12 Chemistry chapter 6 explains the applications of various metals in our day-to-day life. Different terminologies such as refining, calcination, concentration, benefaction, roasting, etc. are very well explained here. The concepts of thermodynamics for the extraction of copper, aluminium and zinc are also mentioned here.

This chapter contains the details of various chemical processes associated with metallurgy. Being an important chapter, it will help the students score well during the examinations.

NCERT IN-TEXT QUESTIONS

Question 1.
Name some ores which can be concentrated by magnetic separation method.
Answer:
Only those ores can be concentrated by magnetic separation method in which either the ore particles or the impurities associated with it are of magnetic nature. For example, ores of iron such haematite (Fe₂O₃), magnetite (Fe₃O₄), siderite (FeCO₃) are magnetic and can be concentrated by this method. Similarly, casseterite (SnO₂) an ore of tin is non-magnetic while the impurities of tungstates of iron and chromium are of magnetic nature. Magnetic separation is effective in this case also.

Question 2.
What is the significance of leaching in the extraction of aluminium?
Answer:
Aluminium contains silica (SiO₂), iron oxide (Fe₂O₃) and titanium oxide (TiO₄) as impurities. These impurities can be removed by the process of leaching. During leaching, the powdered bauxite ore is heated with a concentrated (45%) solution of NaOH at 473-523 K, where alumina dissolves as sodium meta-aluminate and silica as sodium silicate leaving Fe₂O₃, TiO₂ and other impurities behind:

The impurities are filtered off and solution of sodium meta-aluminate is neutralised by passing CO₂ when hydrated alumina separates out while sodium silicate remains in solution. The hydrated aluminathus obtained is filtered, dried and heated to give back pure alumina.

Thus, by leaching, pure alumina can be obtained from bauxite ore.

Question 3.
The reaction : Cr₂O₂(s) + 2Al(s) → Al₂O₃(s) + 2Cr(s) ; ∆G° = – 421 kJ is thermodynamically feasible as is apparent from the value of ∆G°. Why does not it take place at room temperature ?
Answer:
Though the reaction is feasible, it does not proceed at room temperature because all the reactants and products are solids. At elevated temperature, when chromium starts melting, the reaction becomes feasible.

Question 4
Is it true that under certain conditions, Mg can reduce Al₂O₃ and Al can reduce MgO ? What are those conditions ?
Answer:
If we look at the Ellingham diagram, it becomes evident that the plots for Al and Mg cross each other at 1350 °C (1623 K). Below this temperature, Mg can reduce Al₂O₃ and above this temperature, Al can reduce MgO.

NCERT EXERCISE

Question 1.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
Answer:
Copper is a comparatively less active metal as its reduction potential i.e. E° (Cu²⁺/Cu) is high (+0-34V). It can be displaced from a solution of Cu²⁺ ions by more active metals which have E° value lower than copper. For example, E° of zinc (Zn²⁺/Zn) is -0.76V and thus, zinc can displace copper from the solution of Cu²⁺ ions. In contrast, to displace zinc from a solution of Zn²⁺ ions a more reactive metal than zinc is required like, Na, K, Mg, Ca, etc. But, the more active metals readily react with water forming their corresponding ions and evolve hydrogen gas.
[2Na + 2H₂O → 2NaOH + H₂],
Thus, it is difficult to displace zinc from a solution of Zn²⁺ ions. Elance, copper can be extracted by hydrometallurgy but not zinc.

Question 2.
What is the role of the depressant in the froth floatation process?
Answer:
A depressant suppresses the formation of froth with a particular compound in the froth floatation process by reacting chemically with it. Thus, it helps in the separation of two metal sulphides present together in a particular ore.

In actual process, the sulphide ore is finely powdered and is mixed with water to form a slurry in a tank as shown in Fig. 6.3. To this oily component of the sulphide ore particles by water. As a result, ore and oil constitute hydrophobic or water-repelling components while gangue and water form a lyophilic or water-attracting component.

Question 3.
Why is the extraction of copper from its sulphide ore difficult than that from its oxide through reduction?
Answer:
Sulphide ore of copper (Cu₂S) cannot be directly reduced by either coke or hydrogen because ∆_fG° of Cu₂S is more than those of CS₂ and H₂S that will be formed as a result of the reaction.
These reactions are, therefore, not feasible. However, the ∆_fG° of Cu₂O is lower than that of CO₂. Therefore, the sulphide ore is first roasted to Cu₂O which is then reduced.

Question 4.
Explain
(i) zone refining
(ii) column chromatography. (C.B.S.E. Sample Paper 2015)
Answer:
(i) Zone refining (Fractional Crystallisation). This method is used only if a metal in almost pure state is required. Metals like germanium and gallium which are used in semiconductors are purified by this method. The principle of zone refining is based on the fact that impurities are more soluble in molten metal than in solid metal.

(ii) Chromatographic method. This method can also be employed on small scale for the purification of certain metals. Adsorption chromatography is normally used for this purpose. Different components present in a given sample are adsorbed to a different extent on the surface of the adsorbent.

Question 5.
Out of C and CO, which is a better reducing agent at 673 K?
Answer:
When carbon reacts with dioxygen, two reactions are possible
C_(s) + O_2(g) → CO_2(g) ……(i)
2C_(s) + O_2(g) → 2CO_(g) …..(ii)
When CO is used as a reducing agent, it gets oxidized to CO₂
2CO + O₂ → 2CO₂ …(iii)
It is clear from the Ellingham diagram that at 673K the ∆G° for the oxidation of CO to CO₂ is more negative than the reaction (i) and reaction (ii). Therefore, CO is a better reducing agent than C. It is supported by the fact that the curve for the reaction (iii) lies below the curve for the reaction (i) and reaction (ii) at 673K. An element below in the Ellingham diagram reduces the oxide of other metal which lies above it.

Question 6.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?
Answer:
The common elements present in the anode mud are antimony, selenium, tellurium, silver, gold, and platinum. These elements settle down under anode as anode mud because they are less reactive and are not effected by CuSO₄ – H₂SO₄solution.

Question 7.
Write the chemical reactions which take place in different zones in the blast furnace during the extraction of iron.
Answer:

The chemical reactions which take place in the blast furnace are briefly discussed.
Zone of combustion. At the bottom of the furnace, the blast of hot air causes the combustion of coke into carbon dioxide. The reaction is highly exothermic and a temperature of nearly 2170 K develops. It supplies most of the heat required for the process.

Zone of heat absorption. As CO₂ gas rises up, it combines with more coke to form carbon monoxide. Since the reaction is endothermic, the temperature in the middle of the furnace is nearly 1570 K. It further decreases as the reaction proceeds.

Zone of slag formation. In the middle of the furnace, the temperature is nearly 1123 K. Here limestone (CaCO₃) decomposes to form CaO and CO₂. The former (CaO) combines with silica (SiO₂) which is an impurity in the haematite ore to form calcium silicate (CaSiO₃) that is fusible. This implies that CaO has acted as a basic flux. It has combined with the acidic impurity of Si02 to form slag.

Zone of reduction: This is the upper part of the furnace. The temperature ranges between 500K to 900K. Here haematite (Fe₂O₃) is reduced to FeO.

Further reduction of FeO to Fe occurs at higher temperatures (1123 K) by CO gas.

The direct reduction of iron ore left unreacted also occurs with carbon above 1123 K

Question 8.
Write the chemical reactions which take place in the extraction of zinc from zinc blende.
Answer:
Zinc blende is chemically zinc sulphide (ZnS). It undergoes the following reactions:

Zinc metal is in crude or impure form. It can be refined with the help of electro-refining. In this method, impure zinc is made anode while a plate of pure metal acts as the cathode. The electrolyte is aqueous zinc sulphate containing a small amount of dilute H₂SO₄. On passing electric current, Zn²⁺ ions from the electrolyte migrate towards the cathode and are reduced to zinc metal which gets deposited on the cathode.

From anode, an equivalent amount of zinc gets oxidised to Zn²⁺ ions which migrate to the electrolyte

Question 9.
State the role of silica in the metallurgy of copper.
Answer:
During roasting, copper pyrites are converted into a mixture of FeO and Cu₂0. Thus, acidic flux silica is added during smelting to remove FeO (basic). FeO combines with SiO₂to form famous silicate (FeSiO₃) slag which floats over molten matte.

Question 10.
What is meant by the term chromatography?
Answer:
The term chromatography was originally derived from the Greek word chroma meaning colour and graphy meaning writing because the method was first used for the separation of coloured substances (plant pigments) into individual components. Now the term chromatography has lost its original meaning and the method is widely used for separation, purification and characterization of the components of a mixture whether coloured or colourless.

Question 11.
What criterion is followed for the selection of the stationary phase in chromatography?
Answer:
The stationary phase is selected in such a way that it is capable of adsorbing the impurities more strongly than the elements to be purified. Under this condition, the impurities are retained by stationary phase i. e„ these cannot be eluted easily while the pure component, which is weakly adsorbed is easily eluted.

Question 12.
Describe a method for the refining of nickel.
Answer:
Nickel is refined by Mond’s process.
Mond’s process. Mond’s process is used for the refining of nickel metal. In the process, impure metal is heated in a current of carbon monoxide (CO) at 330 to 350 K to form nickel carbonyl which is of volatile nature. The vapours of the metal carbonyl escape leaving behind impurities. Upon heating to about 450 K, nickel carbonyl decomposes to give pure nickel.

From the above discussion, we may conclude that the basic requirements for the refining of metal by Vand Alkel process and Mond’s process are:

• The metal should form a volatile compound with the available reagent.
• The volatile compound should be easily decomposable so that metal in pure form may be regenerated.

Question 13.
How can you separate alumina from silica in bauxite ore associated with silica? Give equations if any.
Answer:
The purification of bauxite containing silica as the main impurity is done by Serpeck’s process. The powdered ore is mixed with coke and heated to about 2073 K in an atmosphere of nitrogen. Silica (SiO₂) is reduced to silicon which being volatile escapes. Alumina (Al₂O₃) is converted into aluminium nitride (AIN) by reacting with nitrogen. It is hydrolysed upon heating with water to get the precipitate of Al(OH)₃. From the precipitate, Al₂O₃ is recovered upon strong heating.

Question 14.
Giving examples, differentiate between calcination and roasting.
Answer:

Question 15.
How is ‘cast iron’ different from ‘pig iron’?
Answer:
Iron obtained from the blast furnace is called pig iron. It contains about 4% carbon and other impurities of S, P, Si, Mn etc. When pig iron is mixed with scrap iron and coke and then heated in a blast of hot air, some impurities are removed. Cast iron is obtained. It contains 3% carbon and some other impurities. It is hard and brittle.

Question 16.
Differentiate between mineral and ore.
Answer:
The naturally occurring chemical substances in form of which the metal occurs in the earth along with impurities are called minerals. The minerals from which the metal can be extracted conveniently and economically are called ore. Thus, all ores are minerals but all minerals are not ores. For example, iron is found in the earth’s crust oxides, carbonates and sulphides. Out of these minerals of iron, the oxides of iron are employed for the extraction of the metal. Therefore, oxides of iron are the ores of iron. Similarly, aluminium occurs in the earth’s crust in form of two minerals, i.e, bauxite (Al₂O₃. x H₂O) and clay (Al₂O₃.2SiO. 2H₂O). Out of these two minerals, Al can be conveniently and economically extracted from bauxite the ore of alluminium.

Question 17.
Why is copper matte put in the silicon-lined converter?
Answer:
Copper matte consists of a mixture of Cu₂S and Cu₂O. Along with that, it also contains a small amount of FeS and FeO. It is put in a silicon-lined converter known as a Bessemer converter. Some silica (SiO₂) is also added and a blast of hot air is blown. As a result, a number of reactions take place.

Question 18.
What is the role of cryolite in the metallurgy of aluminium?
Answer:
The role of cryolite is two-fold

• It makes alumina a good conductor of electricity.
• It lowers the fusion (melting point) temperature of both from 2323K to about 1140K.

Question 19.
How is leaching carried out in case of low-grade copper?
Answer:
Leaching in case of low-grade copper is carried out by reacting with an acid like H₂SO₄ in the presence of air when copper is oxidised to Cu²⁺ ions which pass into the solution. For example.
2 Cu(s) + 2 H₂SO₄ (aq) + O₂ (g) → 2CuSO₄ (aq) + 2 H₂O(l)
or Cu + 2H⁺ (aq) + 1/2 O₂(g) → Cu²⁺ (aq) + H₂O(l)

Question 20.
Why is zinc not extracted from zinc oxide through reduction using CO?
Answer:
The standard free energy of formation (Δ_f G°) of CO₂ from CO (fig 6.8) is higher than that of the formation-of ZnO from Zn. Therefore, CO cannot be used to reduce ZnO to Zn.

Question 21.
The value of ∆_fG° for the formation of Cr₂O₃ is – 540 kJ mol^-1 and that of Al₂O₃ is – 827 kJ mol^-1. Is the reduction of Cr₂O₃ with Al possible? (Pb. Board2009)
Answer:
The two thermochemical equations may be written as follows :

Since ∆G° comes out to be negative, the reaction is feasible.

Question 22.
Out of C and CO, which is a better reducing agent for ZnO?
Answer:
The Δ_f G° of CO2 from CO is always a higher reduction of ZnO to Zn. In contrast, Δ_f G° of CO from C is lower at a temperature above 1180K while that of CO₂ from C is lower at temperatures above 1270K than Δ_f G° of ZnO. Thus above 1270K, ZnO can be reduced to Zn by C. In actual practice, the reduction is usually carried out around 1673K. Thus out of C and CO, C is a better reducing agent than CO for ZnO.

Question 23.
The choice of a reducing agent in a particular case depends on the thermodynamic factors. How do you agree with this statement? Support your opinion with two examples.
Answer:
Thermodynamic factors have a major role in selecting the reducing agent for a particular reaction.
(i) Only that reagent will be preferred which will lead to decrease in free energy (∆G°) at a certain specific temperature.
(ii) A metal oxide placed lower in the Ellingham diagram cannot be reduced by the metal involved in the formation of the oxide placed higher in the diagram.
Examples:
(0 Al₂O₃ cannot be reduced by Cr present in Cr₂O₃ since the curve for Al₂O₃ is placed below that of Cr₂O₃ in the Ellingham diagram
(ii) CO cannot reduce ZnO because there is hardly any change in free energy (∆G°) as a result of the reaction.

Question 24.
Name the processes from which chlorine is obtained as the by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis?
Answer:
Chlorine is obtained as the by-product in the manufacture of sodium by Down’s process in which molten sodium chloride is subjected to electrolysis.

Sodium obtained by this method is almost pure while chlorine is the by-product.
Chlorine can also be obtained by carrying out the electrolysis of an aqueous solution of sodium chloride. The process is carried in Nelson’s cell. The various reactions which take place are as follows :

At cathode. Both Na⁺ and H⁺ ions migrate towards the cathode but H⁺ ions are discharged in preference to Na⁺ ions since their discharge potential is less. Na⁺ ions remain in the solution.

At anode: Both Cl^– and OH^– ions migrate towards the anode but Cl^– ions are discharged in preference to OH” since their discharge potential is less. OH^– ions remain in the solution.

Thus, in the electrolysis of an aqueous NaCl solution, H₂ gas is evolved at the cathode and chlorine at the anode. The solution contains NaOH and is, therefore, basic in nature.
It is always better to prepare chlorine by Down’s process.

Question 25.
What is the role of a graphite rod in the electrometallurgy of aluminium?
Answer:
In the electrometallurgy of aluminium, oxygen gas is evolved at the anode. It reacts with graphite or carbon (graphite electrodes) to form carbon monoxide and carbon dioxide. In case some other metal electrodes act as an anode, then oxygen will react with aluminium formed during the process to form aluminium oxide (Al₂O₃) which will pass into the reaction mixture. Since graphite is cheaper than aluminium, its wastage or consumption can be tolerated.

Question 26.
Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining
Answer:
(i) 1. This method is based on the principle that the impurities are more soluble in the molten state of metal (the melt) than in the solid-state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves the molten zone of the rod also moves with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then the end with impurities is cut off. Silicon, boron, gallium, indium, etc can be purified by this process.

2. Column chromatography is a technique used to separate components of a mixture where components are in minute quantities. In chromatography, there are two phases: the mobile phase and the stationary phase. The stationary phase is immobile and immiscible. Al₂O₃ column is usually used as the stationary phase in column chromatography. The mobile phase may be a gas, liquid, or supercritical fluid in which the sample extracts dissolve. Then the mobile phase is forced to wave through the stationary phase. The component that is more strongly absorbed on the column takes a long time to travel than the component weakly, absorbed.

(ii) Electrolytic refining is the process of refining impure metal by using electricity. In this process, impure metal is made the anode and a strip of pure metal is made the cathode. A solution of a soluble salt of the same metal is taken as the electrolyte. When an electric current is passed, metal ions from the electrolyte are deposited at the cathode as pure metal and the impure metal from the anode dissolves into the electrolyte in the form options. The impurities present in the impure metal gets collected below the anode. This is called anode mud.

Anode : M → Mⁿ⁺+ ne^–
Cathode : Mⁿ⁺ + ne^– → M.

(iii) Vapour phase refining is the process of refining metal by converting it into its volatile compound and then, decomposing is to obtain a pure metal to carry out this process.
(a) The metal should form a volatile compound with available reagent, and
(b) The volatile compound should be easily decomposable so that the metal can be easily recovered.
Nickel, Zirconium, and titanium are refined using this method.

Question 27.
Predict the conditions under which aluminium can be expected to reduce magnesium oxide.
Answer:
The equations for the formation of the two oxides are :

If we look at the plots for the formation of the two oxides on the Ellingham diagram, we find that they intersect at a certain point. The corresponding value of ∆G° becomes zero for the reduction of MgO by Al metal.

This means that the reduction of MgO by A1 metal can occur below this temperature.
Aluminium (Al) metal can reduce MgO to Mg above this temperature because ∆_fG for Al₂O₃ is less as compared to that of MgO.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 30
We hope the NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements, drop a comment below and we will get back to you at the earliest.