Check the below NCERT MCQ Questions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry with Answers Pdf free download. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. We have provided Some Basic Concepts of Chemistry Class 11 Chemistry MCQs Questions with Answers to help students understand the concept very well.

## Some Basic Concepts of Chemistry Class 11 MCQs Questions with Answers

Some Basic Concepts Of Chemistry MCQ Question 1.
Approximate atomic weight of an element is 26.89. If its equivalent weight is 8.9, the exact atomic weight of element would be
(a) 26.89
(b) 8.9
(c) 17.8
(d) 26.7

Explanation:
Atomic weight = (Equivalent weight × Valency)
=(8.9 × 3) = 26.7
(Valency = (26.89)/(8.9) ≈ 3).

Class 11 Chemistry Chapter 1 MCQ Questions With Answers Question 2.
The number of moles present in 6 gms of carbon is:
(a) 2
(b) 0.5
(c) 5
(d) 1

Explanation:
The molar mass of
12C is 12.0 gmol-1.
12C atoms have a mass of 12.0 g.
Given that: – 6.0 g.
Thus (6.0 g)/ (12.0 gmol-1) = 0.50 mol.

Class 11 Chemistry Chapter 1 MCQ Question 3.
What is the concentration of nitrate ions if equal volumes of 0.1 M AgNO3 and 0.1 M NaCl are mixed together
(a) 0.1 N
(b) 0.2 M
(c) 0.05 M
(d) 0.25 M

Explanation:
0.1 M AgNOA will react with 0.1 M NaCl to form 0.1 M NaNOA.
But as the volume is doubled, conc. of NO3 = $$\frac { (0.1) }{ (2) }$$
= 0.05 M

Some Basic Concepts Of Chemistry Class 11 MCQ With Answers Question 4.
The -ve charged particles is called:
(a) Anion
(b) Cation
(d) Atom

Explanation:
A charged particle, also called an ion, is an atom with a positive or negative charge.
This happens whenever something called an ionic bond forms.
Two particles that have different numbers of electrons (the smallest particle in an atom which is negative) start reacting to each other.
The particle that has the greater amount of electrons takes the other particles electrons.
One becomes positive because it lost an electron, and the other negative because it got another electron.
The two particles become attracted to each other and mix together, making a new kind of particle.

MCQ Questions For Class 11 Chemistry Chapter 1 Question 5.
Which is not a unit of pressure:
(a) Bar
(b) N/m²
(c) Kg/m²
(d) Torr

Explanation:
Pressure is force per unit area so Kg/m² cannot be a unit of pressure.

Some Basic Concepts Of Chemistry Class 11 MCQ Question 6.
What is the normality of a 1 M solution of H3PO4
(a) 0.5 N
(b) 1.0 N
(c) 2.0 N
(d) 3.0 N

Explanation:
H3PO4 is tribasic
So N = 3M
= 3 × 1 = 3.

Chemistry Class 11 Chapter 1 MCQs Question 7.
The total number of ions present in 111 g of CaCl2 is
(a) One Mole
(b) Two Mole
(c) Three Mole
(d) Four Mole

Explanation:
Molecular weight of
CaCl2 = 111g/mol
Ions in one calcium chloride molecule = Ca+2 + 2Cl = 3 ions
Now no. of molecules in 111g/mol of CaCl2 = Avogadros number
= 6.02 × 1023 molecules
So number of ions in 111g/mol of CaCl2
= 3 × 6.02 × 1023 ions
= 3 moles.

Ch 1 Chemistry Class 11 MCQ Question 8.
Which of the following weighs the most?
(a) One g – atom of nitrogen
(b) One mole of water
(c) One mole of sodium
(d) One molecule of H2SO4

Answer: (c) One mole of sodium

Chapter 1 Chemistry Class 11 MCQ Question 9.
Under similar conditions of pressure and temperature, 40 ml of slightly moist hydrogen chloride gas is mixed with 20 ml of ammonia gas, the final volume of gas at the same temperature and pressure will be
(a) 100 ml
(b) 20 ml
(c) 40 ml
(d) 60 ml

Explanation:
NH3(g) + HCl(g)→ NH4Cl(s)
t = 0 40ml 0
t = t 20ml solid
Final volume = 20ml.

Chemistry Class 11 Chapter 1 MCQ Question 10.
An organic compound contains carbon , hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be
(a) CHO
(b) CH4O
(c) CH3O
(d) CH2O

Explanation:

Class 11 Chemistry Ch 1 MCQ Question 11.
Which of the following cannot give iodometric titrations
(a) Fe3+
(b) Cu2+
(c) Pb2+
(d) Ag+

Explanation:
Atom in highest oxidation state can oxidize iodide to liberate I2 which is volumetrically measured by iodometric titration using hypo.
2I →I2
Pb+2 → Lowest oxidation state cannot oxidise iodide to I2

MCQ Class 11 Chemistry Chapter 1 Question 12.
What is the concentration of nitrate ions if equal volumes of 0.1 MAgNO3 and 0.1 M NaCl are mixed together
(a) 0.1 M
(b) 0.2 M
(c) 0.05 M
(d) 0.25 M

Explanation:
0.1M AgNO3 will react with 0.1M NaCl to form 0.1M NaNO3.
But as the volume doubled, conc. of NO2
= 0.12
= 0.05 M

MCQ Of Chapter 1 Chemistry Class 11 Question 13.
The number of moles present in 6 gms of carbon is:
(a) 2
(b) 0.5
(c) 5
(d) 1

Explanation:
The molar mass of
12C is 12.0 gmol-1.
NA (Avogadros number = 6.022 × 1023)
12C atoms have a mass of 12.0 g.
Given that: – 6.0 g.
Thus (6.0 g)/ (12.0 gmol-1) = 0.50 mol.

MCQ On Some Basic Concepts Of Chemistry Question 14.
Which of the following contains same number of carbon atoms as are in 6.0 g of carbon (C – 12)?
(a) 6.0 g Ethane
(b) 8.0 g Methane
(c) 21.0 g Propane
(d) 28.0 g CO

Explanation:
6g carbon
Moles of carbon = (6/12) = 0.5 mol
Number of carbon atoms
= 0.5 × NA =0.5NA (NA is Avogadro number)
6g ethane (C2H6 two atoms of C per mole)
Moles = (6/30) = 0.2 mol
Number of carbon atoms = 0.2 × 2 × NA = 0.4 NA
(Number of carbon atoms = moles of compound × number of C atoms per mol × Avogadro number)
8g methane (CH4)
Moles = (8/16) = 0.5 mol
Number of carbon atoms = 0.5 × 1 × NA = 0.5 NA
21g propane (C3H8)
Moles = (21/44) =0.48 mol
Number of carbon atoms = 0.48 × 3 × NA = 1.44 NA
28g CO
Moles = (28/28) =1 mol
Number of carbon atoms = 1 × 1 × NA = NA

MCQ Of Ch 1 Chemistry Class 11 Question 15.
The significant figures in 3400 are
(a) 2
(b) 5
(c) 6
(d) 4

Explanation:
As we know that all non-zero unit are significant number.

Question 16.
A symbol not only represents the name of the element but also represents
(a) Atomic Mass
(b) Atomic Number
(c) Atomicity
(d) Atomic Volume

Explanation:
Symbol of an element represents the name along with the number of same atoms in room condition, which is its atomicity, of that element.

Question 17.
What is the normality of a 1 M solution of H3PO4
(a) 0.5 N
(b) 1.0 N
(c) 2.0 N
(d) 3.0 N

Explanation:
H3PO4 is tribasic
So N = 3M
= 3 × 1 = 3.

Question 18.
Which of the following weighs the most?
(a) One g – atom of nitrogen
(b) One mole of water
(c) One mole of sodium
(d) One molecule of H2SO4

Answer: (c) One mole of sodium

Question 19.
The sulphate of a metal M contains 9.87% of M. This sulphate is isomorphous with ZnSO4.7H2O. The atomic weight of M is
(a) 40.3
(b) 36.3
(c) 24.3
(d) 11.3

Explanation:
As the given sulphate is isomorphous with ZnSO4.7H2O its formula would be MSO4.7H2O.m is the atomic weight of M, molecular weight of MSO4.7H2O
= (m + 32 + 64 + 126) = (m + 222)
Hence % of M= [(m)/ (m+222)] × 100 = 9.87(given)
Or
100 m = (9.87 m + 222 × 9.87)
Or 90.13 m = (222 × 9.87)
Or m = (222 × 9.87)/(90.13)
= 24.3.

Question 20.
The S.I unit of temperature is :
(a) Kelvin
(b) Celsius
(c) Fahrenheit