JEE Chemistry Solutions MCQs Set 20

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Chapter-wise Objective Questions: Solutions

View or download the dedicated Solutions MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Chemistry evaluations.

Multiple Choice Questions

Question 1: 9.8 g of H2SO4 is present in 2 litres of a solution. The molarity of the solution is
(a) 0.1 M
(b) 0.05 M
(c) 0.2 M
(d) 0.01 M
Show Answer & Explanation

Answer: (b) 0.05 M

Explanation:
1. Moles of H2SO4 = 9.8/98 = 0.1 mol.
2. Molarity = 0.1 mol / 2 L = 0.05 M.

Teacher's Note:
1. Molarity is moles per litre of solution, not per litre of solvent.

Question 2: What is the molarity of a solution containing 5 g of sodium hydroxide in 250 mL of solution?
(a) 0.5
(b) 1.0
(c) 2.0
(d) 0.1
Show Answer & Explanation

Answer: (a) 0.5

Explanation:
1. Moles of NaOH = 5/40 = 0.125 mol.
2. Molarity = 0.125/0.250 = 0.5 M.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 3: The normality of 0.3 M phosphorous acid (H3PO3) is
(a) 0.1
(b) 0.9
(c) 0.3
(d) 0.6
Show Answer & Explanation

Answer: (d) 0.6

Explanation:
1. H3PO3 is dibasic: only two of its three hydrogen atoms are acidic.
2. N = 0.3 × 2 = 0.6 N.

Teacher's Note:
1. Count only the replaceable (acidic) hydrogens when finding basicity.

Question 4: 20 mL of HCl solution requires 19.85 mL of 0.01 M NaOH solution for complete neutralisation. The molarity of the HCl solution is
(a) 0.0099
(b) 0.099
(c) 0.99
(d) 9.9
Show Answer & Explanation

Answer: (a) 0.0099

Explanation:
1. M1V1 = M2V2 for a 1 : 1 reaction.
2. M(HCl) = 0.01 × 19.85/20 = 0.0099 M.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 5: If 5.85 g of NaCl (molecular weight 58.5) is dissolved in water and the solution is made up to 0.5 litre, the molarity of the solution will be
(a) 0.2
(b) 0.4
(c) 1.0
(d) 0.1
Show Answer & Explanation

Answer: (a) 0.2

Explanation:
1. Moles of NaCl = 5.85/58.5 = 0.1 mol.
2. Molarity = 0.1/0.5 = 0.2 M.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 6: A mixture has 18 g water and 414 g ethanol. The mole fraction of water in the mixture is (assume ideal behaviour)
(a) 0.1
(b) 0.4
(c) 0.7
(d) 0.9
Show Answer & Explanation

Answer: (a) 0.1

Explanation:
1. Moles of water = 18/18 = 1; moles of ethanol = 414/46 = 9.
2. Mole fraction of water = 1/(1 + 9) = 0.1.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 7: 36 g of water and 828 g of ethyl alcohol form an ideal solution. The mole fraction of water in it is
(a) 1.0
(b) 0.7
(c) 0.4
(d) 0.1
Show Answer & Explanation

Answer: (d) 0.1

Explanation:
1. Moles of water = 36/18 = 2; moles of ethanol = 828/46 = 18.
2. x(water) = 2/20 = 0.1.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 8: If 1 M and 2.5 litre NaOH solution is mixed with another 0.5 M and 3 litre NaOH solution, the molarity of the resulting solution will be
(a) 1.0 M
(b) 0.73 M
(c) 0.50 M
(d) 0.80 M
Show Answer & Explanation

Answer: (b) 0.73 M

Explanation:
1. Total moles = 1 × 2.5 + 0.5 × 3 = 4.0 mol.
2. Total volume = 5.5 L, so M = 4.0/5.5 = 0.73 M.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 9: 3.0 molal NaOH solution has a density of 1.110 g/mL. The molarity of the solution is
(a) 3.0504
(b) 3.64
(c) 3.05
(d) 2.9732
Show Answer & Explanation

Answer: (d) 2.9732

Explanation:
1. Mass of solution containing 3 mol NaOH in 1000 g water = 1000 + 120 = 1120 g.
2. Volume = 1120/1.110 = 1009 mL.
3. Molarity = 3/1.009 = 2.97 M.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 10: Two solutions of a non-electrolyte are mixed: 480 mL of 1.5 M first solution and 520 mL of 1.2 M second solution. What is the molarity of the final mixture?
(a) 1.20 M
(b) 1.50 M
(c) 1.344 M
(d) 2.70 M
Show Answer & Explanation

Answer: (c) 1.344 M

Explanation:
1. Moles = 1.5 × 0.480 + 1.2 × 0.520 = 0.720 + 0.624 = 1.344 mol.
2. Volume = 1.000 L, so M = 1.344 M.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 11: The mole fraction of water in a 20% aqueous solution of H2O2 is
(a) 77/68
(b) 68/77
(c) 20/80
(d) 80/20
Show Answer & Explanation

Answer: (b) 68/77

Explanation:
1. In 100 g: H2O2 = 20/34 = 10/17 mol; water = 80/18 = 40/9 mol.
2. x(water) = (40/9)/(40/9 + 10/17) = 680/770 = 68/77.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 12: 10 mL of concentrated H2SO4 (18 M) is diluted to 1 litre. The approximate strength of the dilute acid is
(a) 0.18 N
(b) 0.09 N
(c) 0.36 N
(d) 1800 N
Show Answer & Explanation

Answer: (c) 0.36 N

Explanation:
1. M1V1 = M2V2: 18 × 10 = M2 × 1000, so M2 = 0.18 M.
2. N = 2 × 0.18 = 0.36 N.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 13: The normality of a 10 volume hydrogen peroxide solution is
(a) 0.176
(b) 3.52
(c) 1.78
(d) 0.88
Show Answer & Explanation

Answer: (c) 1.78

Explanation:
1. 10 volume means 1 L of solution gives 10 L of O2 at STP.
2. Equivalent weight of H2O2 corresponds to 5.6 L of O2, so N = 10/5.6 = 1.78 N.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 14: The volume strength of 1.5 N H2O2 solution is
(a) 4.8
(b) 5.2
(c) 8.8
(d) 8.4
Show Answer & Explanation

Answer: (d) 8.4

Explanation:
1. Volume strength = 5.6 × normality = 5.6 × 1.5 = 8.4.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 15: In a mixture of 1 g H2 and 8 g O2, the mole fraction of hydrogen is
(a) 0.667
(b) 0.5
(c) 0.33
(d) None of these
Show Answer & Explanation

Answer: (a) 0.667

Explanation:
1. Moles of H2 = 1/2 = 0.5; moles of O2 = 8/32 = 0.25.
2. x(H2) = 0.5/0.75 = 0.667.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 16: The molarity of pure water is
(a) 55.6
(b) 5.56
(c) 100
(d) 18
Show Answer & Explanation

Answer: (a) 55.6

Explanation:
1. 1 L of water weighs 1000 g = 1000/18 = 55.6 mol.
2. Molarity = 55.6 M.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 17: What is the molality of a solution having 18 g of glucose (molecular weight 180) dissolved in 500 g of water?
(a) 1 m
(b) 0.5 m
(c) 0.2 m
(d) 2 m
Show Answer & Explanation

Answer: (c) 0.2 m

Explanation:
1. Moles of glucose = 18/180 = 0.1 mol.
2. Molality = 0.1 mol / 0.5 kg = 0.2 m.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 18: A solution of Al2(SO4)3 (d = 1.253 g/mL) contains 22% salt by weight. The molarity, normality and molality of the solution are respectively
(a) 0.805 M, 4.83 N, 0.825 m
(b) 0.825 M, 4.83 N, 0.805 m
(c) 4.83 M, 0.825 N, 0.805 m
(d) None of these
Show Answer & Explanation

Answer: (a) 0.805 M, 4.83 N, 0.825 m

Explanation:
1. Molar mass of Al2(SO4)3 = 342 g/mol. M = (1000 × 1.253 × 0.22)/342 = 0.806 M.
2. Valence factor = 6, so N = 6 × 0.806 = 4.83 N.
3. Molality = (22/342)/0.078 kg = 0.825 m.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 19: Calculate the molality of a 1 litre solution of 93% H2SO4 (weight/volume). The density of the solution is 1.84 g/mL.
(a) 10.43
(b) 20.36
(c) 12.05
(d) 14.05
Show Answer & Explanation

Answer: (a) 10.43

Explanation:
1. 1 L has mass 1840 g and contains 930 g H2SO4 (93% w/v).
2. Water = 1840 − 930 = 910 g; moles of H2SO4 = 930/98 = 9.49.
3. Molality = 9.49/0.910 = 10.43 m.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 20: Concentrated H2SO4 has a density of 1.98 g/mL and is 98% H2SO4 by weight. Its normality is
(a) 2 N
(b) 19.8 N
(c) 39.6 N
(d) 98 N
Show Answer & Explanation

Answer: (c) 39.6 N

Explanation:
1. Molarity = (1980 × 0.98)/98 = 19.8 M.
2. H2SO4 is dibasic, so N = 2 × 19.8 = 39.6 N.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 21: What is the molarity of an HCl solution which contains 49% by weight of solute and whose specific gravity is 1.41?
(a) 15.25
(b) 16.75
(c) 18.92
(d) 20.08
Show Answer & Explanation

Answer: (c) 18.92

Explanation:
1. Mass of 1 L = 1410 g; HCl = 0.49 × 1410 = 690.9 g.
2. Moles = 690.9/36.5 = 18.93, so the molarity is about 18.92 M.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 22: The mole fraction of the solute in a one molal aqueous solution is
(a) 0.027
(b) 0.036
(c) 0.018
(d) 0.009
Show Answer & Explanation

Answer: (c) 0.018

Explanation:
1. 1 molal means 1 mol solute in 1000 g (55.5 mol) of water.
2. x(solute) = 1/(1 + 55.5) = 0.0177 ≈ 0.018.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

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