JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 40

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Chapter-wise Objective Questions: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: Which of the following contains the maximum number of atoms?
(a) 4 g of H2
(b) 16 g of O2
(c) 28 g of N2
(d) 18 g of H2O
Show Answer & Explanation

Answer: (a) 4 g of H2

Explanation:
1. Moles of atoms: H2 = 2 × 2 = 4; O2 = 0.5 × 2 = 1; N2 = 1 × 2 = 2; H2O = 1 × 3 = 3.
2. 4 g of H2 has the most atoms.
3. Answer: (a)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 2: 5.6 L of oxygen at STP contains:
(a) 6.02 × 1023 atoms
(b) 3.01 × 1023 atoms
(c) 1.505 × 1023 atoms
(d) 0.7525 × 1023 atoms
Show Answer & Explanation

Answer: (b) 3.01 × 1023 atoms

Explanation:
1. 5.6/22.4 = 0.25 mol O2 = 0.5 mol of atoms.
2. Atoms = 0.5 × 6.02 × 1023 = 3.01 × 1023.
3. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 3: The sum of the number of protons, electrons and neutrons in 12 g of 126C is:
(a) 1.8
(b) 12.044 × 1023
(c) 1.084 × 1025
(d) 10.84 × 1023
Show Answer & Explanation

Answer: (c) 1.084 × 1025

Explanation:
1. 12 g of 12C is 1 mol = 6.022 × 1023 atoms.
2. Each atom has 6 + 6 + 6 = 18 particles: 18 × 6.022 × 1023 = 1.084 × 1025.
3. Answer: (c)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 4: Which of the following has the highest mass?
(a) 1 g atom of C
(b) 1/2 mole of CH4
(c) 10 mL of water
(d) 3.011 × 1023 atoms of oxygen
Show Answer & Explanation

Answer: (a) 1 g atom of C

Explanation:
1. 1 g atom of C = 12 g.
2. ½ mol CH4 = 8 g; 10 mL of water = 10 g; 3.011 × 1023 O atoms = 0.5 mol = 8 g.
3. So 1 g atom of C is the heaviest.
4. Answer: (a)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 5: Which of the following contains the least number of molecules?
(a) 4.4 g CO2
(b) 3.4 g NH3
(c) 1.6 g CH4
(d) 3.2 g SO2
Show Answer & Explanation

Answer: (d) 3.2 g SO2

Explanation:
1. Moles: CO2 0.1; NH3 0.2; CH4 0.1; SO2 3.2/64 = 0.05.
2. SO2 has the fewest molecules.
3. Answer: (d)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 6: 4.4 g of an unknown gas occupies 2.24 L of volume at STP. The gas may be:
(a) N2O
(b) CO
(c) CO2
(d) Both N2O and CO2
Show Answer & Explanation

Answer: (d) Both N2O and CO2

Explanation:
1. Molar mass = 4.4/(2.24/22.4) = 44 g mol−1.
2. N2O and CO2 both have M = 44.
3. Answer: (d)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 7: Which contains the least number of molecules?
(a) 1 g CO2
(b) 1 g N2
(c) 1 g O2
(d) 1 g H2
Show Answer & Explanation

Answer: (a) 1 g CO2

Explanation:
1. For equal mass, the number of molecules is inversely proportional to molar mass.
2. CO2 has the largest molar mass (44), so the fewest molecules.
3. Answer: (a)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 8: If V mL of the vapours of a substance at NTP weigh W g, then the molecular weight of the substance is:
(a) (W/V) × 22400
(b) (V/W) × 22.4
(c) (W − V) × 22400
(d) (W × 1)/(V × 22400)
Show Answer & Explanation

Answer: (a) (W/V) × 22400

Explanation:
1. V mL weighs W g, so 22400 mL weighs W × 22400/V g.
2. M = (W/V) × 22400.
3. Answer: (a)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 9: If 3.01 × 1020 molecules are removed from 98 mg of H2SO4, then the number of moles of H2SO4 left are:
(a) 0.1 × 10−3
(b) 0.5 × 10−3
(c) 1.66 × 10−3
(d) 9.95 × 10−2
Show Answer & Explanation

Answer: (b) 0.5 × 10−3

Explanation:
1. 98 mg = 10−3 mol.
2. Removed = 3.01 × 1020/6.02 × 1023 = 0.5 × 10−3 mol.
3. Left = 0.5 × 10−3 mol.
4. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 10: The vapour density of a gas is 11.2. The volume occupied by 2.4 g of this gas at STP will be:
(a) 11.2 L
(b) 2.24 L
(c) 22.4 L
(d) 2.4 L
Show Answer & Explanation

Answer: (d) 2.4 L

Explanation:
1. M = 2 × VD = 22.4 g mol−1.
2. Moles = 2.4/22.4; volume = (2.4/22.4) × 22.4 = 2.4 L.
3. Answer: (d)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 11: A person adds 1.71 g of sugar (C12H22O11) to sweeten his tea. The number of carbon atoms added is (molar mass of sugar = 342):
(a) 3.6 × 1022
(b) 7.2 × 1021
(c) 0.05
(d) 6.6 × 1022
Show Answer & Explanation

Answer: (a) 3.6 × 1022

Explanation:
1. Moles of sugar = 1.71/342 = 0.005.
2. Carbon atoms = 0.005 × 12 × 6.02 × 1023 = 3.6 × 1022.
3. Answer: (a)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 12: An oxide of metal M has 40% by mass of oxygen. The atomic mass of M is 24. The empirical formula of the oxide is:
(a) M2O
(b) M2O3
(c) MO
(d) M3O4
Show Answer & Explanation

Answer: (c) MO

Explanation:
1. M : O = 60/24 : 40/16 = 2.5 : 2.5 = 1 : 1.
2. Empirical formula MO.
3. Answer: (c)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 13: A compound contains 38.8% C, 16.0% H and 45.2% N. The formula of the compound would be:
(a) CH3NH2
(b) CH3CN
(c) C2H5CN
(d) CH2(NH)2
Show Answer & Explanation

Answer: (a) CH3NH2

Explanation:
1. C : H : N = 38.8/12 : 16/1 : 45.2/14 = 3.23 : 16 : 3.23 = 1 : 5 : 1.
2. CH5N = CH3NH2.
3. Answer: (a)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 14: 2.2 g of a compound of phosphorus and sulphur has 1.24 g of P in it. Its empirical formula is:
(a) P2S3
(b) P3S2
(c) P3S4
(d) P4S3
Show Answer & Explanation

Answer: (d) P4S3

Explanation:
1. Sulphur = 2.2 − 1.24 = 0.96 g.
2. P : S = 1.24/31 : 0.96/32 = 0.04 : 0.03 = 4 : 3, so P4S3.
3. Answer: (d)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 15: The numbers of atoms of Cr and O are 4.8 × 1010 and 9.6 × 1010 respectively. Its empirical formula is:
(a) Cr2O3
(b) CrO2
(c) Cr3O4
(d) CrO5
Show Answer & Explanation

Answer: (b) CrO2

Explanation:
1. Cr : O = 4.8 : 9.6 = 1 : 2.
2. Empirical formula CrO2.
3. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 16: 9 g of Al will react with (2Al + (3/2)O2 → Al2O3):
(a) 6 g O2
(b) 8 g O2
(c) 9 g O2
(d) 4 g O2
Show Answer & Explanation

Answer: (b) 8 g O2

Explanation:
1. Moles of Al = 9/27 = 1/3.
2. O2 needed = (1/3) × (3/4) = 0.25 mol = 8 g.
3. Answer: (b)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 17: The equation 2Al(s) + (3/2)O2(g) → Al2O3(s) shows that:
(a) 2 mol of Al reacts with 3/2 mol of O2 to produce 7/2 mol of Al2O3
(b) 2 g of Al reacts with 3/2 g of O2 to produce one mole of Al2O3
(c) 2 g of Al reacts with 3/2 litre of O2 to produce 1 mol of Al2O3
(d) 2 mol of Al reacts with 3/2 mol of O2 to produce 1 mol of Al2O3
Show Answer & Explanation

Answer: (d) 2 mol of Al reacts with 3/2 mol of O2 to produce 1 mol of Al2O3

Explanation:
1. The coefficients give mole ratios: 2 mol Al + 1.5 mol O2 → 1 mol Al2O3.
2. Answer: (d)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 18: The number of litres of air (20% oxygen by volume) required to burn 8 litres of C2H2 is approximately:
(a) 40
(b) 60
(c) 80
(d) 100
Show Answer & Explanation

Answer: (d) 100

Explanation:
1. C2H2 + 2.5O2 → 2CO2 + H2O: 8 L needs 20 L of O2.
2. Air = 20/0.20 = 100 L.
3. Answer: (d)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 19: If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is (3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl):
(a) 0.7
(b) 0.5
(c) 0.3
(d) 0.1
Show Answer & Explanation

Answer: (d) 0.1

Explanation:
1. Compare mole/coefficient: BaCl2 0.5/3 = 0.167; Na3PO4 0.2/2 = 0.1.
2. Na3PO4 is limiting, so 0.1 mol Ba3(PO4)2 forms.
3. Answer: (d)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

Question 20: If 8 mL of uncombined O2 remain after exploding O2 with 4 mL of hydrogen, the number of mL of O2 originally present was:
(a) 12
(b) 2
(c) 10
(d) 4
Show Answer & Explanation

Answer: (c) 10

Explanation:
1. 2H2 + O2 → 2H2O: 4 mL H2 uses 2 mL O2.
2. Original O2 = 8 + 2 = 10 mL.
3. Answer: (c)

Teacher's Note:
1. Convert masses to moles first.
2. Check units and the stoichiometric ratio.

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