JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 03

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Multiple Choice Questions

Question 1: The number of atoms present in 16 g of oxygen is
(a) 6.02 × 1011.5
(b) 3.01 × 1023
(c) 3.01 × 1011.5
(d) 6.02 × 1023
Show Answer & Explanation

Answer: (d) 6.02 × 1023

Explanation:
1. 16 g of oxygen atoms = 16/16 = 1 mol of atoms.
2. Number of atoms = 1 × 6.02 × 1023.

Teacher's Note:
1. Always ask whether the question gives a mass of atoms or of molecules.
2. Oxygen written as 'oxygen' without a formula is treated here as atoms.

Question 2: The number of atoms in 4.25 g of NH3 is approximately
(a) 1 × 1023
(b) 1.5 × 1023
(c) 2 × 1023
(d) 6 × 1023
Show Answer & Explanation

Answer: (d) 6 × 1023

Explanation:
1. Moles of NH3 = 4.25/17 = 0.25 mol.
2. Atoms per molecule = 4, so atoms = 0.25 × 4 × 6.02 × 1023 = 6 × 1023.

Teacher's Note:
1. Multiply the number of molecules by the atomicity to get atoms.
2. NH3 has four atoms per molecule.

Question 3: Which of the following contains the maximum number of oxygen atoms?
(a) 1 g of O
(b) 1 g of O2
(c) 1 g of O3
(d) All have the same number of atoms
Show Answer & Explanation

Answer: (d) All have the same number of atoms

Explanation:
1. Each sample contains only oxygen atoms, so 1 g of any of them has 1/16 mol of atoms.
2. The number of atoms is therefore the same in all three.

Teacher's Note:
1. Same mass of the same element means the same number of atoms, whatever the molecular form.
2. Do not be misled by the formula O2 or O3.

Question 4: The number of atoms present in 0.5 g atom of nitrogen is the same as the atoms in
(a) 12 g of C
(b) 32 g of S
(c) 8 g of oxygen
(d) 24 g of Mg
Show Answer & Explanation

Answer: (c) 8 g of oxygen

Explanation:
1. 0.5 g atom of N = 0.5 mol of atoms.
2. 8 g of O = 8/16 = 0.5 mol of atoms; 12 g C = 1 mol, 32 g S = 1 mol, 24 g Mg = 1 mol.

Teacher's Note:
1. One gram atom is one mole of atoms.
2. Convert every option to moles before comparing.

Question 5: 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 = 4/2 × 2 = 4; O2 = 16/32 × 2 = 1; N2 = 28/28 × 2 = 2; H2O = 18/18 × 3 = 3.
2. The maximum is 4 mol of atoms in 4 g of H2.

Teacher's Note:
1. Light molecules give a large number of moles for the same mass.
2. Remember to multiply the moles by atomicity.

Question 6: The number of neutrons present in 1.7 g of ammonia is
(a) NA
(b) NA/10 × 4
(c) (NA/10) × 7
(d) NA × 10 × 7
Show Answer & Explanation

Answer: (c) (NA/10) × 7

Explanation:
1. Moles of NH3 = 1.7/17 = 0.1 mol.
2. Neutrons per molecule: N has 7, H has 0, so 7 neutrons.
3. Neutrons = 0.1 × 7 × NA = (NA/10) × 7.

Teacher's Note:
1. Neutrons = mass number minus atomic number for each atom.
2. Ordinary hydrogen has no neutrons.

Question 7: 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. Moles of O2 = 5.6/22.4 = 0.25 mol.
2. Atoms = 0.25 × 2 × 6.02 × 1023 = 3.01 × 1023.

Teacher's Note:
1. One mole of any gas occupies 22.4 L at STP.
2. Multiply by 2 because O2 is diatomic.

Question 8: The number of oxygen atoms in 8 g of ozone is
(a) 6.02 × 1023
(b) \( \frac{6.02 \times 10^{23}}{2} \)
(c) \( \frac{6.02 \times 10^{23}}{3} \)
(d) \( \frac{6.02 \times 10^{23}}{6} \)
Show Answer & Explanation

Answer: (b) \( \frac{6.02 \times 10^{23}}{2} \)

Explanation:
1. 8 g of O3 contains 8 g of oxygen atoms = 8/16 = 0.5 mol.
2. Number of atoms = 0.5 × 6.02 × 1023 = \( \frac{6.02 \times 10^{23}}{2} \).

Teacher's Note:
1. Mass of oxygen atoms equals the mass of ozone given.
2. Use atomic mass 16, not molecular mass 48, for atoms.

Question 9: The number of atoms in n moles of a gas can be given by
(a) n × Avogadro number × atomicity
(b) \( \frac{n \times \text{Avogadro number}}{\text{atomicity}} \)
(c) \( \frac{\text{Avogadro number} \times \text{atomicity}}{n} \)
(d) None of these
Show Answer & Explanation

Answer: (a) n × Avogadro number × atomicity

Explanation:
1. Number of molecules = n × NA.
2. Each molecule has 'atomicity' atoms, so atoms = n × NA × atomicity.

Teacher's Note:
1. Atoms = molecules × atomicity.
2. Check the formula with 1 mol of O2, which has 2 NA atoms.

Question 10: The sum of the number of protons, electrons and neutrons in 12 g of \( {}^{12}_{6}\text{C} \) 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 carbon-12 = 1 mol = 6.022 × 1023 atoms.
2. Each atom has 6 protons, 6 electrons and 6 neutrons = 18 particles.
3. Total = 18 × 6.022 × 1023 = 1.084 × 1025.

Teacher's Note:
1. Count particles per atom first and then scale by Avogadro's number.
2. Neutrons = 12 − 6 = 6.

Question 11: The weight of one atom of uranium is 238 amu. Its actual weight (in g) is
(a) 1.43 × 1026
(b) 3.94 × 10-22
(c) 6.99 × 10-23
(d) 1.53 × 10-22
Show Answer & Explanation

Answer: (b) 3.94 × 10-22

Explanation:
1. 1 amu = 1.66 × 10-24 g.
2. Actual weight = 238 × 1.66 × 10-24 = 3.95 × 10-22 g, which matches 3.94 × 10-22.

Teacher's Note:
1. Actual mass of an atom = atomic mass in amu × 1.66 × 10-24 g.
2. Equivalently divide the molar mass by NA.

Question 12: The actual weight of a molecule of water is
(a) 18 g
(b) 2.99 × 10-23 g
(c) Both (1) and (2) are correct
(d) 1.66 × 10-24 g
Show Answer & Explanation

Answer: (b) 2.99 × 10-23 g

Explanation:
1. Molar mass of water = 18 g, so one molecule = 18/(6.022 × 1023) = 2.99 × 10-23 g.
2. 18 g is the mass of one mole, not of one molecule.

Teacher's Note:
1. Mass of one molecule = molar mass divided by NA.
2. Do not confuse molar mass with molecular mass in grams.

Question 13: What is the mass of a molecule of CH4?
(a) 16 g
(b) 26.6 × 1022 g
(c) 2.66 × 10-23 g
(d) 16 NA g
Show Answer & Explanation

Answer: (c) 2.66 × 10-23 g

Explanation:
1. Mass of one molecule = 16/(6.02 × 1023) = 2.66 × 10-23 g.

Teacher's Note:
1. Divide the molar mass by NA for one molecule.
2. The answer is tiny, around 10-23 g for small molecules.

Question 14: Which of the following has the highest mass?
(a) 1 g atom of C
(b) 1/2 mol 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. 1/2 mol CH4 = 8 g; 10 mL water = 10 g (density 1 g/mL); 3.011 × 1023 O atoms = 0.5 mol = 8 g.
3. The highest mass is 12 g.

Teacher's Note:
1. Convert all options to grams.
2. A gram atom means the mass of one mole of atoms.

Question 15: 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 = 4.4/44 = 0.1; NH3 = 3.4/17 = 0.2; CH4 = 1.6/16 = 0.1; SO2 = 3.2/64 = 0.05.
2. The least is 0.05 mol, in SO2.

Teacher's Note:
1. Molecules are proportional to moles.
2. Heavy molecules give fewer moles for a given mass.

Question 16: The number of molecules in 4.25 g of NH3 is
(a) 1.505 × 1023
(b) 3.01 × 1023
(c) 6.02 × 1023
(d) None of these
Show Answer & Explanation

Answer: (a) 1.505 × 1023

Explanation:
1. Moles = 4.25/17 = 0.25 mol.
2. Molecules = 0.25 × 6.02 × 1023 = 1.505 × 1023.

Teacher's Note:
1. Compare with question 2: atoms were four times the molecules.
2. Always state whether the count is of atoms or molecules.

Question 17: Elements A and B form two compounds B2A3 and B2A. 0.05 mol of B2A3 weighs 9.0 g and 0.10 mol of B2A weighs 10 g. The atomic weights of A and B are
(a) 20 and 30
(b) 30 and 40
(c) 40 and 30
(d) 30 and 20
Show Answer & Explanation

Answer: (c) 40 and 30

Explanation:
1. Molar mass of B2A3 = 9.0/0.05 = 180; molar mass of B2A = 10/0.10 = 100.
2. 2B + 3A = 180 and 2B + A = 100; subtracting gives 2A = 80, so A = 40.
3. Then 2B = 60, so B = 30.

Teacher's Note:
1. Set up two equations from two molar masses.
2. Subtract to eliminate the common term.

Question 18: 5.6 L of oxygen at NTP is equivalent to
(a) 1 mol
(b) 1/2 mol
(c) 1/4 mol
(d) 1/8 mol
Show Answer & Explanation

Answer: (c) 1/4 mol

Explanation:
1. Moles = 5.6/22.4 = 1/4.

Teacher's Note:
1. Molar volume at NTP is 22.4 L.
2. 5.6 L is one quarter of that.

Question 19: 4.4 g of an unknown gas occupies 2.24 L at STP. The gas may be
(a) N2O
(b) CO
(c) CO2
(d) Both (1) and (3)
Show Answer & Explanation

Answer: (d) Both (1) and (3)

Explanation:
1. Moles = 2.24/22.4 = 0.1 mol, so molar mass = 4.4/0.1 = 44 g/mol.
2. N2O = 44 and CO2 = 44, while CO = 28.
3. So the gas may be N2O or CO2.

Teacher's Note:
1. Different gases can have the same molar mass.
2. Use the molar mass only to narrow down the options.

Question 20: 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 masses, moles are inversely proportional to molar mass.
2. CO2 has the largest molar mass (44), so 1 g of it has the fewest molecules.

Teacher's Note:
1. Equal masses: the heavier molecule means fewer molecules.
2. No calculation is needed when comparing equal masses.

Question 21: If V mL of the vapours of a substance at NTP weigh W g, the molecular weight of the substance is
(a) \( \frac{W}{V} \times 22400 \)
(b) \( \frac{V}{W} \times 22.4 \)
(c) \( (W - V) \times 22400 \)
(d) \( \frac{W \times 1}{V \times 22400} \)
Show Answer & Explanation

Answer: (a) \( \frac{W}{V} \times 22400 \)

Explanation:
1. V mL of vapour is the fraction V/22400 of a mole.
2. Molar mass = W ÷ (V/22400) = (W/V) × 22400.

Teacher's Note:
1. Molar mass = mass of 22400 mL of the gas.
2. Check the units: g per mL times mL per mol gives g/mol.

Question 22: If 3.01 × 1020 molecules are removed from 98 mg of H2SO4, the number of moles of H2SO4 left is
(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 = 0.098 g = 0.001 mol = 1 × 10-3 mol.
2. Moles removed = 3.01 × 1020/6.02 × 1023 = 0.5 × 10-3.
3. Moles left = 1 × 10-3 − 0.5 × 10-3 = 0.5 × 10-3.

Teacher's Note:
1. Convert mg to g carefully.
2. Subtract the moles removed from the initial moles.

Question 23: A gas is found to have the formula (CO)_x. Its vapour density is 70. The value of x must be
(a) 7
(b) 4
(c) 5
(d) 6
Show Answer & Explanation

Answer: (c) 5

Explanation:
1. Molar mass = 2 × vapour density = 140.
2. Mass of CO = 28, so x = 140/28 = 5.

Teacher's Note:
1. Molar mass = 2 × V.D.
2. The formula unit CO has mass 28.

Question 24: 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. Molar mass = 2 × 11.2 = 22.4 g/mol.
2. 2.4 g = 2.4/22.4 mol, and volume = (2.4/22.4) × 22.4 = 2.4 L.

Teacher's Note:
1. The molar mass in grams numerically equals the volume of one mole in litres here.
2. Always double the V.D. first.

Question 25: The volume of a gas in a discharge tube is 1.12 × 10-7 mL at STP. The number of molecules of the gas in the tube is
(a) 3.01 × 104
(b) 3.01 × 1015
(c) 3.01 × 1012
(d) 3.01 × 1016
Show Answer & Explanation

Answer: (c) 3.01 × 1012

Explanation:
1. Moles = 1.12 × 10-7/22400 = 5 × 10-12 mol.
2. Molecules = 5 × 10-12 × 6.02 × 1023 = 3.01 × 1012.

Teacher's Note:
1. Use 22400 mL as the molar volume when the volume is in mL.
2. Keep track of powers of ten.

Question 26: 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 mol.
2. Carbon atoms = 0.005 × 12 × 6.02 × 1023 = 3.6 × 1022.

Teacher's Note:
1. Multiply by the number of carbon atoms in one molecule.
2. 0.05 is a number of moles, not atoms.

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