JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 34

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Chapter-wise Objective Questions: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: The density (g mL-1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 by mass is (H2SO4 = 98 g mol-1)
(a) 1.22
(b) 1.45
(c) 1.64
(d) 1.88
Show Answer & Explanation

Answer: (a) 1.22

Explanation:
1. M = 10 d × (mass %) / molar mass, so d = 3.60 × 98 / (10 × 29).
2. d = 352.8/290 = 1.22 g mL-1.

Teacher's Note:
1. Rearrange the molarity formula for density.
2. Check: 1 L has 1220 g, with 354 g acid, which is 29%.

Question 2: A 5.2 molal aqueous solution of methyl alcohol is supplied. The mole fraction of methyl alcohol in the solution is
(a) 0.086
(b) 0.190
(c) 0.050
(d) 0.100
Show Answer & Explanation

Answer: (a) 0.086

Explanation:
1. 1000 g of water = 55.5 mol, with 5.2 mol CH3OH.
2. Mole fraction = 5.2/(5.2 + 55.5) = 0.086.

Teacher's Note:
1. Molal means moles per kg of solvent.
2. 1000 g of water is 55.5 mol.

Question 3: The molality of a solution of 0.0100 g of urea, (NH2)2CO, in 0.3000 dm3 of water is (urea = 60 g mol-1)
(a) 3.33 × 10-2 m
(b) 0.555 m
(c) 33.3 m
(d) 5.55 × 10-4 m
Show Answer & Explanation

Answer: (d) 5.55 × 10-4 m

Explanation:
1. Moles of urea = 0.0100/60 = 1.667 × 10-4 mol.
2. Mass of water = 0.300 kg, so molality = 1.667 × 10-4/0.300 = 5.55 × 10-4 m.

Teacher's Note:
1. 0.3 dm3 of water has a mass of 0.3 kg.
2. Divide moles by the mass of solvent in kg.

Question 4: The molarity of the solution obtained by mixing 750 mL of 0.5 M HCl with 250 mL of 2 M HCl is
(a) 1.75 M
(b) 1.00 M
(c) 0.975 M
(d) 0.875 M
Show Answer & Explanation

Answer: (d) 0.875 M

Explanation:
1. Moles = 0.75 × 0.5 + 0.25 × 2 = 0.375 + 0.5 = 0.875 mol.
2. Volume = 1 L, so molarity = 0.875 M.

Teacher's Note:
1. Add moles, add volumes.
2. The result must lie between 0.5 M and 2 M.

Question 5: The molecular formula of a resin used for water softening is C8H7SO3Na (mass 206). The maximum uptake of Ca2+ by the resin, in mol per gram of resin, is
(a) 1/206
(b) 2/309
(c) 1/412
(d) 1/103
Show Answer & Explanation

Answer: (c) 1/412

Explanation:
1. Two resin units (2 Na+) exchange for one Ca2+.
2. Moles of resin per gram = 1/206, so Ca2+ = 1/(2 × 206) = 1/412 mol per gram.

Teacher's Note:
1. Charge balance: one Ca2+ replaces two Na+.
2. Per gram of resin means divide by the molar mass.

Question 6: 1 g of a carbonate M2CO3 treated with excess HCl produces 0.01186 mol of CO2. The molar mass of M2CO3 in g mol-1 is
(a) 1186
(b) 118.6
(c) 84.3
(d) 11.86
Show Answer & Explanation

Answer: (c) 84.3

Explanation:
1. One mole of M2CO3 gives one mole of CO2.
2. Molar mass = 1/0.01186 = 84.3 g mol-1.

Teacher's Note:
1. Mass divided by moles gives the molar mass.
2. The 1 : 1 ratio is the key.

Question 7: The vapour density of carbon dioxide is
(a) 12
(b) 44
(c) 32
(d) 22
Show Answer & Explanation

Answer: (d) 22

Explanation:
1. Vapour density = molar mass / 2 = 44/2 = 22.

Teacher's Note:
1. Vapour density is half the molar mass.
2. It has no unit.

Question 8: Molality of a solution is the number of
(a) moles of solute per 100 g of solvent
(b) gram equivalents of solute per kilogram of solvent
(c) gram moles of solute per 1000 cm3 of solution
(d) moles of solute per 1000 g of solvent
Show Answer & Explanation

Answer: (d) moles of solute per 1000 g of solvent

Explanation:
1. Molality = moles of solute per kilogram (1000 g) of solvent.

Teacher's Note:
1. Molarity uses litres of solution, molality uses kg of solvent.

Question 9: A (atomic mass 12.01) and B (35.5) combine to form X, with 4 mol of B reacting with 1 mol of A to give 1 mol of X. The mass of one mole of X is
(a) 74.0 g
(b) 47.5 g
(c) 154.0 g
(d) 166.0 g
Show Answer & Explanation

Answer: (c) 154.0 g

Explanation:
1. X is AB4: 12.01 + 4 × 35.5 = 154.01 g mol-1.

Teacher's Note:
1. The mole ratio gives the formula.

Question 10: If 1 dm3 of a gas weighs 2.5 g at STP, its gram molecular mass is
(a) 224 g
(b) 56 g
(c) 11.2 g
(d) 22.4 g
Show Answer & Explanation

Answer: (b) 56 g

Explanation:
1. Mass of 22.4 dm3 = 2.5 × 22.4 = 56 g, which is the molar mass.

Teacher's Note:
1. Molar mass = density at STP × 22.4.

Question 11: How many moles of air are in the lungs of an average adult with a lung capacity of 3.8 L at 1.0 atm and body temperature 37 °C? (R = 0.0821 L atm K-1 mol-1)
(a) 0.25 mol
(b) 0.15 mol
(c) 1.15 mol
(d) 2.25 mol
Show Answer & Explanation

Answer: (b) 0.15 mol

Explanation:
1. n = PV/RT = (1.0 × 3.8)/(0.0821 × 310) = 0.149 mol, about 0.15 mol.

Teacher's Note:
1. Use T in kelvin: 37 + 273 = 310 K.

Question 12: A saline solution for rinsing contact lenses is made by dissolving 400 mg of NaCl in water and diluting to 100 mL. Its molarity is (Na = 23, Cl = 35.5)
(a) 0.09564 M
(b) 0.0684 M
(c) 1.0684 M
(d) 0.00684 M
Show Answer & Explanation

Answer: (b) 0.0684 M

Explanation:
1. Moles = 0.400/58.5 = 6.84 × 10-3 mol.
2. Molarity = 6.84 × 10-3/0.100 L = 0.0684 M.

Teacher's Note:
1. Convert mg to g first.
2. Volume in litres is 0.1.

Question 13: Cystine has 26.7% sulphur by mass. If its molecule contains two sulphur atoms, its molecular mass is
(a) 2400
(b) 120
(c) 24
(d) 240
Show Answer & Explanation

Answer: (d) 240

Explanation:
1. Mass of 2 S atoms = 64 u.
2. Molecular mass = 64/0.267 = 240 u.

Teacher's Note:
1. Mass of the known atoms divided by their mass fraction.

Question 14: One gram mole of CO2 contains
(a) 6.022 × 1023 atoms of carbon
(b) 3 gram atoms of CO2
(c) 3.011 × 1023 molecules of CO2
(d) 6.022 × 1023 atoms of oxygen
Show Answer & Explanation

Answer: (a) 6.022 × 1023 atoms of carbon

Explanation:
1. 1 mol CO2 = 6.022 × 1023 molecules, each with one C atom, so 6.022 × 1023 C atoms and twice as many O atoms.

Teacher's Note:
1. Atoms of an element = molecules × atoms per molecule.

Question 15: Which of the following solutions are unimolar? (A) 0.46 g C2H5OH in 10 mL (B) 110.98 g CaCl2 in 1000 mL (C) 0.23 g CH3OH in 100 mL (D) 5.88 g NaCl in 1000 mL
(a) B and C
(b) C and D
(c) A and D
(d) A and B
Show Answer & Explanation

Answer: (d) A and B

Explanation:
1. (A) 0.46/46 = 0.01 mol in 0.010 L = 1 M. (B) 110.98/110.98 = 1 mol in 1 L = 1 M.
2. (C) 0.23/32 = 0.0072 mol in 0.1 L = 0.072 M. (D) 5.88/58.5 = 0.1 mol per L = 0.1 M.
3. So (A) and (B) are unimolar.

Teacher's Note:
1. Convert every sample to mol L-1.
2. Unimolar means 1 M.

Question 16: The maximum amount of CH3Cl that can be obtained by reacting 20.0 g of CH4 with 10.0 g of Cl2 (CH4 + Cl2 → CH3Cl + HCl) is
(a) 63.1 g
(b) 31.6 g
(c) 30.0 g
(d) 7.1 g
Show Answer & Explanation

Answer: (d) 7.1 g

Explanation:
1. CH4 = 1.25 mol; Cl2 = 10/71 = 0.141 mol, so Cl2 is limiting.
2. CH3Cl = 0.141 mol × 50.5 = 7.1 g.

Teacher's Note:
1. Identify the limiting reagent.
2. Molar mass of CH3Cl is 50.5 g.

Question 17: A mixture of aluminium and zinc weighing 1.67 g was dissolved in acid and evolved 1.69 L of hydrogen at 273 K and 1 atm. The amount of aluminium in the mixture is approximately
(a) 2.2 g
(b) 2.0 g
(c) 1.8 g
(d) 1.2 g
Show Answer & Explanation

Answer: (d) 1.2 g

Explanation:
1. H2 = 1.69/22.4 = 0.0754 mol. Let Al = a g and Zn = 1.67 - a g.
2. Al gives 3/2 mol H2 per mol and Zn gives 1 mol: 1.5a/27 + (1.67 - a)/65.4 = 0.0754.
3. Solving gives a = 1.24 g, about 1.2 g.

Teacher's Note:
1. Write the hydrogen balance with one unknown.
2. The total mass cannot be less than the Al mass, which rules out larger options.

Question 18: The largest number of molecules is present in 1 g of
(a) H2O
(b) N2O5
(c) C2H5OH
(d) CO2
Show Answer & Explanation

Answer: (a) H2O

Explanation:
1. For equal mass, the number of molecules is largest for the smallest molar mass.
2. Molar masses: CO2 44, H2O 18, C2H5OH 46, N2O5 108, so H2O wins.

Teacher's Note:
1. Number of molecules = mass / molar mass × N_A.

Question 19: An ammonia solution is labelled density 0.91 g cm-3, 25% w/w. The molarity is
(a) 17 M
(b) 11.5 M
(c) 13.4 M
(d) 15 M
Show Answer & Explanation

Answer: (c) 13.4 M

Explanation:
1. M = 10 × d × % / molar mass = 10 × 0.91 × 25/17 = 13.4 M.

Teacher's Note:
1. Use M = 10 d (%)/M_2.

Question 20: 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
(a) 0.5
(b) 0.1
(c) 0.7
(d) 0.2
Show Answer & Explanation

Answer: (b) 0.1

Explanation:
1. 3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl.
2. BaCl2 allows 0.5/3 = 0.167 mol and Na3PO4 allows 0.2/2 = 0.1 mol, so Na3PO4 is limiting: 0.1 mol.

Teacher's Note:
1. Compare moles / coefficient.

Question 21: The total number of electrons present in 8.0 g of methane is
(a) 3.01 × 1024
(b) 4.8 × 1025
(c) 4.8 × 1024
(d) 3.01 × 1023
Show Answer & Explanation

Answer: (a) 3.01 × 1024

Explanation:
1. 8.0 g CH4 = 0.5 mol; each molecule has 6 + 4 = 10 electrons.
2. Electrons = 0.5 × 10 × 6.02 × 1023 = 3.01 × 1024.

Teacher's Note:
1. Add the atomic numbers of all atoms in the molecule.

Question 22: 1000 mL of a gas weighs 1.5 g at NTP. Its gram molecular mass is
(a) 11.2 g
(b) 22.4 g
(c) 15 g
(d) 33.6 g
Show Answer & Explanation

Answer: (d) 33.6 g

Explanation:
1. Mass of 22.4 L = 1.5 × 22.4 = 33.6 g.

Teacher's Note:
1. Molar mass = mass per litre × 22.4.

Question 23: The percentages of C, H and N in an organic compound are 40%, 13.3% and 46.7%. The empirical formula is
(a) CH2N
(b) CH4N
(c) CH5N
(d) C3H9N3
Show Answer & Explanation

Answer: (b) CH4N

Explanation:
1. Moles: C = 3.33, H = 13.3, N = 3.33, so C : H : N = 1 : 4 : 1.

Teacher's Note:
1. Divide by the smallest number of moles.

Question 24: The ideal mass (kg) of aluminium obtained from 1 metric ton of Al2O3 ore (Al = 27, O = 16) is
(a) 1000
(b) 795
(c) 530
(d) 265
Show Answer & Explanation

Answer: (c) 530

Explanation:
1. 1000 kg Al2O3 contains 1000 × 54/102 = 529 kg of Al, about 530 kg.

Teacher's Note:
1. Mass fraction of Al in Al2O3 is 54/102.

Question 25: An opium alkaloid contains 70.8% C, 6.2% H and 4.1% N, the rest being oxygen. Its empirical formula is
(a) C10H11O3N
(b) C20H20N2
(c) C20H21O4N
(d) C21H20N
Show Answer & Explanation

Answer: (c) C20H21O4N

Explanation:
1. Oxygen = 100 - 81.1 = 18.9%.
2. Moles: C = 5.90, H = 6.2, N = 0.293, O = 1.18; dividing by 0.293 gives C : H : N : O = 20 : 21 : 1 : 4.

Teacher's Note:
1. Find the oxygen by difference first.

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