JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 29

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Chapter-wise Objective Questions: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: How many moles of electrons weigh one kilogram? (mass of an electron = \( 9.108 \times 10^{-31} \) kg)
(a) \( \frac{1}{9.108} \times 10^{31} \)
(b) \( \frac{6.023}{9.108} \times 10^{54} \)
(c) \( \frac{1}{9.108 \times 6.023} \times 10^{8} \)
(d) \( 6.023 \times 10^{23} \)
Show Answer & Explanation

Answer: (c) \( \frac{1}{9.108 \times 6.023} \times 10^{8} \)

Explanation:
1. Mass of one mole of electrons = \( 9.108 \times 10^{-31} \times 6.023 \times 10^{23} \) kg.
2. Moles in 1 kg = \( 1/(9.108 \times 6.023) \times 10^{8} \).

Teacher's Note:
1. Moles = total mass / molar mass.
2. Mass of one mole of electrons is about \( 5.5 \times 10^{-7} \) kg.

Question 2: The number of atoms in 560 g of Fe (atomic mass 56 g/mol) is:
(a) twice that in 70 g of N
(b) half that in 20 g of H
(c) both (1) and (2)
(d) none of these
Show Answer & Explanation

Answer: (c) both (1) and (2)

Explanation:
1. 560 g Fe = 10 mol.
2. 70 g N = 5 mol, so Fe has twice as many.
3. 20 g H = 20 mol, so Fe has half as many. Both are correct.

Teacher's Note:
1. Convert all to moles.
2. Compare ratios.

Question 3: Which has the maximum number of atoms?
(a) 108 g of Ag (108)
(b) 24 g of C (12)
(c) 56 g of Fe (56)
(d) 27 g of Al (27)
Show Answer & Explanation

Answer: (b) 24 g of C (12)

Explanation:
1. Moles: C = 2, Fe = 1, Al = 1, Ag = 1.
2. Carbon has the largest number of moles, so the most atoms.

Teacher's Note:
1. Atoms are proportional to moles.
2. Light elements give more atoms per gram.

Question 4: How many moles of magnesium phosphate, Mg\(_3\)(PO\(_4\))\(_2\), contain 0.25 mole of oxygen atoms?
(a) \( 1.25 \times 10^{-2} \)
(b) \( 2.5 \times 10^{-2} \)
(c) \( 3.125 \times 10^{-2} \)
(d) 0.02
Show Answer & Explanation

Answer: (c) \( 3.125 \times 10^{-2} \)

Explanation:
1. One formula unit has 8 oxygen atoms.
2. Moles = 0.25/8 = \( 3.125 \times 10^{-2} \).

Teacher's Note:
1. Count atoms in the formula.
2. Divide by 8.

Question 5: The abundances of the isotopes \(^{54}\)Fe, \(^{56}\)Fe and \(^{57}\)Fe are 5%, 90% and 5% respectively. The atomic mass of Fe is:
(a) 55.75
(b) 55.85
(c) 55.95
(d) 56.05
Show Answer & Explanation

Answer: (c) 55.95

Explanation:
1. Average = 0.05 \(\times\) 54 + 0.90 \(\times\) 56 + 0.05 \(\times\) 57 = 2.7 + 50.4 + 2.85 = 55.95.

Teacher's Note:
1. Weighted average of isotopic masses.
2. Weights are the abundances.

Question 6: Which properties of an element may have non-integral values? (1) Atomic weight (2) Atomic number (3) Atomic volume
(a) 1 and 3 only
(b) 2 and 3 only
(c) 1 and 2 only
(d) 1, 2 and 3
Show Answer & Explanation

Answer: (a) 1 and 3 only

Explanation:
1. Atomic weight is an average over isotopes and can be fractional; atomic volume can also take any value.
2. Atomic number counts protons, so it is always an integer.

Teacher's Note:
1. Average atomic mass is non-integral.
2. Atomic number is a count.

Question 7: Which of the following are equal to \( 10^{-2} \) atm? (1) 0.76 cm Hg (2) 7.6 torr (3) 0.076 dm Hg (4) 0.0076 torr
(a) 1, 2 and 3 only
(b) 1 and 4 only
(c) 2, 3 and 4 only
(d) all four
Show Answer & Explanation

Answer: (a) 1, 2 and 3 only

Explanation:
1. \( 10^{-2} \) atm = 7.6 mm Hg = 7.6 torr = 0.76 cm Hg = 0.076 dm Hg.
2. 0.0076 torr is a thousand times smaller.

Teacher's Note:
1. 1 atm = 760 mm Hg.
2. Convert mm to cm and dm.

Question 8: Oxygen exerts the same pressure under which of the following conditions? (1) 2 L, 27 \(^\circ\)C (2) 1 L, 150 K (3) 4 L, 54 \(^\circ\)C (4) 10 L, 1227 \(^\circ\)C
(a) 1, 2 and 4 only
(b) 1, 2 and 3 only
(c) all four
(d) 3 and 4 only
Show Answer & Explanation

Answer: (a) 1, 2 and 4 only

Explanation:
1. For the same moles, P is proportional to T/V.
2. T/V: 300/2 = 150; 150/1 = 150; 327/4 = 81.75; 1500/10 = 150.
3. So (1), (2) and (4) give the same pressure.

Teacher's Note:
1. Convert \(^\circ\)C to K first.
2. Compare T/V.

Question 9:
Assertion (A): The gram molecular weight of O\(_2\) is 32 g.
Reason (R): The relative atomic weight of oxygen is 32.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. Molar mass of O\(_2\) is 2 \(\times\) 16 = 32 g, so A is true.
2. The relative atomic weight of oxygen is 16, so R is false.

Teacher's Note:
1. Distinguish atomic mass from molecular mass.
2. O\(_2\) is diatomic.

Question 10:
Assertion (A): One mole of all ideal gases exerts the same pressure in the same volume at the same temperature.
Reason (R): The behaviour of ideal gases is independent of their nature.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. From P = nRT/V the pressure depends only on n, T and V.
2. That is because ideal gas behaviour does not depend on the nature of the gas, so R explains A.

Teacher's Note:
1. Ideal gases have no intermolecular forces and negligible molecular volume.
2. Same n, T and V give the same P.

Question 11:
Assertion (A): The value of the universal gas constant depends upon the choice of the system of units.
Reason (R): The values of the universal gas constant are 8.314 J/(mol K), 0.0821 L atm/(mol K) and 2 cal/(mol K).
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R has the same physical meaning but a different numerical value in different units.
2. The different values quoted in R explain why the numerical value depends on the units.

Teacher's Note:
1. R = 8.314 J mol\(^{-1}\) K\(^{-1}\).
2. Use the unit that matches P and V.

Question 12: A vessel of 25 L contains 20 g of an ideal gas X at 300 K. The pressure exerted by the gas is 1 atm. 20 g of ideal gas Y is then added at the same temperature and the total pressure becomes 3 atm. On further addition of 20 g of ideal gas Z, the pressure becomes 7 atm. The ideal gas equation applies to the mixture. Take R = 1/12 L atm K\(^{-1}\) mol\(^{-1}\). Find the molar mass of gas X.
(a) 30 g
(b) 10 g
(c) 5 g
(d) 20 g
Show Answer & Explanation

Answer: (d) 20 g

Explanation:
1. RT = (1/12) \(\times\) 300 = 25 L atm/mol.
2. n(X) = PV/RT = 1 \(\times\) 25/25 = 1 mol.
3. Molar mass = 20 g/1 mol = 20 g/mol.

Teacher's Note:
1. The vessel and temperature stay fixed, so P is proportional to total moles.
2. Each added gas adds moles.

Question 13: A vessel of 25 L contains 20 g of an ideal gas X at 300 K. The pressure exerted by the gas is 1 atm. 20 g of ideal gas Y is then added at the same temperature and the total pressure becomes 3 atm. On further addition of 20 g of ideal gas Z, the pressure becomes 7 atm. The ideal gas equation applies to the mixture. Take R = 1/12 L atm K\(^{-1}\) mol\(^{-1}\). Identify the correct statement(s): I. Gas Y is lighter than gas X. II. Gas Z is lighter than gas Y.
(a) I only
(b) II only
(c) both I and II
(d) none of the statements
Show Answer & Explanation

Answer: (c) both I and II

Explanation:
1. Total moles after Y = 3, so n(Y) = 2 and M(Y) = 20/2 = 10 g/mol.
2. After Z the total is 7, so n(Z) = 4 and M(Z) = 5 g/mol.
3. Since 10 < 20 and 5 < 10, both statements are correct.

Teacher's Note:
1. For equal masses, more moles means lighter molecules.
2. Use pressure increments.

Question 14: A vessel of 25 L contains 20 g of an ideal gas X at 300 K. The pressure exerted by the gas is 1 atm. 20 g of ideal gas Y is then added at the same temperature and the total pressure becomes 3 atm. On further addition of 20 g of ideal gas Z, the pressure becomes 7 atm. The ideal gas equation applies to the mixture. Take R = 1/12 L atm K\(^{-1}\) mol\(^{-1}\). Find the average molar mass of the mixture of gases X, Y and Z.
(a) 40/7
(b) 50/7
(c) 60/7
(d) 20
Show Answer & Explanation

Answer: (c) 60/7

Explanation:
1. Total mass = 60 g.
2. Total moles = 1 + 2 + 4 = 7.
3. Average molar mass = 60/7 g/mol.

Teacher's Note:
1. Average molar mass = total mass / total moles.
2. Moles are proportional to pressure.

Question 15: How much time (in years) would it take to distribute one Avogadro number of wheat grains if \( 10^{10} \) grains are distributed each second?
(a) \( 1.9 \times 10^{9} \) years
(b) \( 6.0 \times 10^{13} \) years
(c) \( 1.9 \times 10^{3} \) years
(d) \( 1.9 \times 10^{6} \) years
Show Answer & Explanation

Answer: (d) \( 1.9 \times 10^{6} \) years

Explanation:
1. Time = \( 6.022 \times 10^{23}/10^{10} = 6.022 \times 10^{13} \) s.
2. One year = \( 3.15 \times 10^{7} \) s.
3. Years = \( 6.022 \times 10^{13}/3.15 \times 10^{7} = 1.9 \times 10^{6} \).

Teacher's Note:
1. Convert seconds to years carefully.
2. This shows how large Avogadro's number is.

Question 16: The weight of one atom of uranium is 238 amu. Its actual weight in grams is:
(a) \( 1.43 \times 10^{26} \) g
(b) \( 3.95 \times 10^{-24} \) g
(c) \( 3.95 \times 10^{-22} \) g
(d) \( 2.38 \times 10^{-22} \) g
Show Answer & Explanation

Answer: (c) \( 3.95 \times 10^{-22} \) g

Explanation:
1. Actual weight = 238 \(\times\) 1.66 \(\times 10^{-24}\) g = \( 3.95 \times 10^{-22} \) g.

Teacher's Note:
1. 1 amu = \( 1.66 \times 10^{-24} \) g.
2. Or divide 238 by N\(_A\).

Question 17: The weight of \( 12.044 \times 10^{23} \) atoms of carbon is:
(a) 12 g
(b) 36 g
(c) 24 g
(d) 6 g
Show Answer & Explanation

Answer: (c) 24 g

Explanation:
1. Moles = \( 12.044 \times 10^{23}/6.022 \times 10^{23} \) = 2.
2. Mass = 2 \(\times\) 12 = 24 g.

Teacher's Note:
1. 2 moles of carbon.
2. Mass = moles \(\times\) 12.

Question 18: How many grams of silicon are present in 35 gram atoms of silicon? (Si = 28)
(a) 1.25 g
(b) 0.8 g
(c) 35 g
(d) 980 g
Show Answer & Explanation

Answer: (d) 980 g

Explanation:
1. Mass = number of gram atoms \(\times\) atomic mass = 35 \(\times\) 28 = 980 g.

Teacher's Note:
1. Gram atom = mole.
2. Mass = moles \(\times\) molar mass.

Question 19: The total number of nucleons present in 12 g of \(^{12}\)C atoms is:
(a) \( 6.022 \times 10^{23} \)
(b) 12
(c) \( 3.6 \times 10^{24} \)
(d) \( 7.23 \times 10^{24} \)
Show Answer & Explanation

Answer: (d) \( 7.23 \times 10^{24} \)

Explanation:
1. 12 g of C-12 = 1 mol = N\(_A\) atoms.
2. Each atom has 12 nucleons, so total = 12 \(\times 6.022 \times 10^{23} = 7.23 \times 10^{24}\).

Teacher's Note:
1. Nucleons = protons + neutrons = mass number.
2. Multiply by N\(_A\).

Question 20: Find the total number of neutrons in 7 mg of \(^{14}\)C.
(a) \( 1.8 \times 10^{21} \)
(b) \( 2.4 \times 10^{21} \)
(c) \( 3.0 \times 10^{20} \)
(d) \( 4.8 \times 10^{21} \)
Show Answer & Explanation

Answer: (b) \( 2.4 \times 10^{21} \)

Explanation:
1. Moles of C-14 = \( 7 \times 10^{-3}/14 = 5 \times 10^{-4} \).
2. Atoms = \( 5 \times 10^{-4} \times 6.022 \times 10^{23} = 3.01 \times 10^{20} \).
3. Neutrons per atom = 14 \(-\) 6 = 8, so total = \( 2.4 \times 10^{21} \).

Teacher's Note:
1. Neutrons = mass number \(-\) atomic number.
2. Multiply atoms by neutrons per atom.

Question 21: How many atoms are there in 100 amu of helium?
(a) 4
(b) 100
(c) 25
(d) 50
Show Answer & Explanation

Answer: (c) 25

Explanation:
1. Mass of one He atom = 4 amu.
2. Atoms in 100 amu = 100/4 = 25.

Teacher's Note:
1. Divide the total mass by the mass of one atom.
2. He is monatomic.

Question 22: The density of liquid mercury is 13.6 g/cm\(^3\). How many moles of mercury are there in 1 litre of the metal? (Hg = 200)
(a) 34
(b) 136
(c) 13.6
(d) 68
Show Answer & Explanation

Answer: (d) 68

Explanation:
1. Mass of 1 L = 1000 cm\(^3\) \(\times\) 13.6 = 13600 g.
2. Moles = 13600/200 = 68.

Teacher's Note:
1. Mass = density \(\times\) volume.
2. 1 L = 1000 cm\(^3\).

Question 23: Chlorine has two isotopes: \(^{35}\)Cl (75%, molar mass 35.0 g) and \(^{37}\)Cl (25%, molar mass 37.0 g). The average atomic mass of chlorine is:
(a) 35.0
(b) 35.5
(c) 36.0
(d) 36.5
Show Answer & Explanation

Answer: (b) 35.5

Explanation:
1. Average = 0.75 \(\times\) 35 + 0.25 \(\times\) 37 = 26.25 + 9.25 = 35.5.

Teacher's Note:
1. Weighted average.
2. This is the familiar 35.5 u.

Question 24: The average atomic mass of magnesium is 24.31 amu. It is composed of 79 mole % \(^{24}\)Mg and the remaining 21 mole % \(^{25}\)Mg and \(^{26}\)Mg. The mole % of \(^{26}\)Mg is:
(a) 5%
(b) 11%
(c) 21%
(d) 10%
Show Answer & Explanation

Answer: (d) 10%

Explanation:
1. Let the mole % of Mg-26 be x. Then Mg-25 is 21 \(-\) x.
2. 24.31 \(\times\) 100 = 24 \(\times\) 79 + 25 \(\times\) (21 \(-\) x) + 26 \(\times\) x, giving 2431 = 1896 + 525 + x.
3. x = 10.

Teacher's Note:
1. Set up the weighted average equation.
2. The three percentages add to 100.

Question 25: A sample of ethane has the same mass as 10.0 million molecules of methane. How many C\(_2\)H\(_6\) molecules does the sample contain?
(a) \( 1.875 \times 10^{7} \)
(b) \( 3.75 \times 10^{7} \)
(c) \( 5.33 \times 10^{6} \)
(d) \( 1.0 \times 10^{7} \)
Show Answer & Explanation

Answer: (c) \( 5.33 \times 10^{6} \)

Explanation:
1. Mass is the same, so numbers are inversely proportional to molar masses.
2. Number of C\(_2\)H\(_6\) molecules = \( 10^7 \times 16/30 = 5.33 \times 10^6 \).

Teacher's Note:
1. The heavier molecule (ethane) gives the smaller number.
2. Number of molecules is inversely proportional to molar mass for equal mass.

Question 26: The number of neutrons in 5 g of D\(_2\)O (D is \(^2_1\)H) is:
(a) \( 6.0 \times 10^{23} \)
(b) \( 1.5 \times 10^{24} \)
(c) \( 3.0 \times 10^{24} \)
(d) \( 1.5 \times 10^{23} \)
Show Answer & Explanation

Answer: (b) \( 1.5 \times 10^{24} \)

Explanation:
1. Molar mass of D\(_2\)O = 20 g, so 5 g = 0.25 mol.
2. Neutrons per molecule: D has 1 each (2 total) and O-16 has 8, so 10.
3. Total = 0.25 \(\times\) 10 \(\times 6.022 \times 10^{23} = 1.5 \times 10^{24}\).

Teacher's Note:
1. Neutrons = mass number \(-\) atomic number for each atom.
2. Add over all atoms in the molecule.

Question 27: If the components of air are N\(_2\) 78%, O\(_2\) 21%, Ar 0.9% and CO\(_2\) 0.1% by volume (or mole), the molecular weight of air is about:
(a) 32.0
(b) 28.0
(c) 28.96
(d) 30.5
Show Answer & Explanation

Answer: (c) 28.96

Explanation:
1. M = 0.78 \(\times\) 28 + 0.21 \(\times\) 32 + 0.009 \(\times\) 40 + 0.001 \(\times\) 44.
2. = 21.84 + 6.72 + 0.36 + 0.044 = 28.96 g/mol.

Teacher's Note:
1. Volume percent equals mole percent for gases.
2. Average molar mass is a weighted average.

Question 28: The weight of 350 mL of a diatomic gas at 0 \(^\circ\)C and 2 atm pressure is 1 g. The weight of one atom of the gas is about:
(a) \( 2.66 \times 10^{-26} \) g
(b) \( 5.3 \times 10^{-23} \) g
(c) \( 2.66 \times 10^{-23} \) g
(d) \( 1.33 \times 10^{-23} \) g
Show Answer & Explanation

Answer: (c) \( 2.66 \times 10^{-23} \) g

Explanation:
1. n = PV/RT = (2 \(\times\) 0.35)/(0.0821 \(\times\) 273) = 0.0312 mol.
2. Molar mass = 1/0.0312 = 32 g/mol, so atomic mass = 16.
3. Mass of one atom = 16/N\(_A\) = \( 2.66 \times 10^{-23} \) g.

Teacher's Note:
1. The gas is oxygen.
2. Divide the atomic mass by N\(_A\).

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