JEE Chemistry Solutions MCQs Set 06

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Multiple Choice Questions

Question 1: 8 g of NaOH is dissolved in one litre of solution. Its molarity is
(a) 0.8 M
(b) 0.4 M
(c) 0.2 M
(d) 0.1 M
Show Answer & Explanation

Answer: (c) 0.2 M

Explanation:
1. Moles of NaOH = 8/40 = 0.2.
2. Molarity = 0.2 mol / 1 L = 0.2 M.

Teacher's Note:
1. Molarity = moles of solute per litre of solution.
2. Molar mass of NaOH is 40 g mol−1.

Question 2: If 18 g of glucose is present in 1000 g of solvent, the solution is said to be
(a) 1 molar
(b) 0.1 molar
(c) 0.5 molar
(d) 0.1 molal
Show Answer & Explanation

Answer: (d) 0.1 molal

Explanation:
1. Moles of glucose = 18/180 = 0.1.
2. Mass of solvent = 1 kg, so molality = 0.1 mol kg−1 = 0.1 molal.

Teacher's Note:
1. Molal refers to moles per kg of solvent.
2. The question gives the mass of solvent, not the volume of solution.

Question 3: The molarity of a solution of sodium chloride (molar mass 58.5 g mol−1) in water containing 5.85 g of sodium chloride in 500 mL of solution is
(a) 0.25
(b) 2.0
(c) 1.0
(d) 0.2
Show Answer & Explanation

Answer: (d) 0.2

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

Teacher's Note:
1. Convert mL to litres before dividing.

Question 4: For preparing a 0.1 M solution of H2SO4 in one litre, we need H2SO4 of mass
(a) 0.98 g
(b) 4.9 g
(c) 49.0 g
(d) 9.8 g
Show Answer & Explanation

Answer: (d) 9.8 g

Explanation:
1. Moles needed = 0.1 × 1 = 0.1 mol.
2. Mass = 0.1 × 98 = 9.8 g.

Teacher's Note:
1. Molar mass of H2SO4 is 98 g mol−1.
2. Mass = molarity × volume × molar mass.

Question 5: The mole fraction of glycerine (C3H5(OH)3) in a solution of 36 g of water and 46 g of glycerine is
(a) 0.46
(b) 0.36
(c) 0.20
(d) 0.40
Show Answer & Explanation

Answer: (c) 0.20

Explanation:
1. Moles of water = 36/18 = 2.
2. Moles of glycerine = 46/92 = 0.5.
3. x = 0.5/(0.5 + 2) = 0.20.

Teacher's Note:
1. Mole fraction = moles of component / total moles.

Question 6: A 1000 g aqueous solution of CaCO3 contains 10 g of calcium carbonate. The concentration of the solution is
(a) 10 ppm
(b) 100 ppm
(c) 1000 ppm
(d) 10,000 ppm
Show Answer & Explanation

Answer: (d) 10,000 ppm

Explanation:
1. ppm = (mass of solute / mass of solution) × 106 = (10/1000) × 106 = 10,000.

Teacher's Note:
1. 1% corresponds to 10,000 ppm.

Question 7: The normality of a 1 M H3PO4 solution is
(a) 0.5 N
(b) 1.0 N
(c) 2.0 N
(d) 3.0 N
Show Answer & Explanation

Answer: (d) 3.0 N

Explanation:
1. N = M × n-factor.
2. H3PO4 is tribasic (n = 3), so N = 3 N.

Teacher's Note:
1. n-factor for acids = basicity.

Question 8: The molarity of a 0.2 N Na2CO3 solution will be
(a) 0.05 M
(b) 0.2 M
(c) 0.1 M
(d) 0.4 M
Show Answer & Explanation

Answer: (c) 0.1 M

Explanation:
1. n-factor of Na2CO3 is 2.
2. M = N/n = 0.2/2 = 0.1 M.

Teacher's Note:
1. M = N / n-factor.

Question 9: The normality of a 0.3 M phosphorous acid (H3PO3) solution is
(a) 0.15
(b) 0.6
(c) 0.9
(d) 0.1
Show Answer & Explanation

Answer: (b) 0.6

Explanation:
1. H3PO3 is dibasic (only two ionisable H), n = 2.
2. N = 0.3 × 2 = 0.6.

Teacher's Note:
1. Phosphorous acid has one P–H bond that does not ionise.

Question 10: The molarity of pure water is
(a) 100 M
(b) 55.5 M
(c) 50 M
(d) 18 M
Show Answer & Explanation

Answer: (b) 55.5 M

Explanation:
1. 1 L of water has about 1000 g, i.e. 1000/18 = 55.5 mol.
2. Molarity = 55.5 M.

Teacher's Note:
1. Density of water is taken as 1 g mL−1.

Question 11: Equal weights of NaCl and KCl are dissolved separately in equal volumes of solution. The molarity of the two solutions will be
(a) equal
(b) that of NaCl will be less than that of KCl
(c) that of NaCl will be more than that of KCl
(d) that of NaCl will be half that of KCl
Show Answer & Explanation

Answer: (c) that of NaCl will be more than that of KCl

Explanation:
1. Molarity ∝ 1/molar mass for equal masses and volumes.
2. NaCl (58.5) has a smaller molar mass than KCl (74.5), so more moles.

Teacher's Note:
1. Same mass, lighter formula unit gives more moles.

Question 12: The mole fraction of oxygen in a mixture of 7 g of nitrogen and 8 g of oxygen is
(a) \( \dfrac{8}{15} \)
(b) 0.5
(c) 0.25
(d) 1.0
Show Answer & Explanation

Answer: (b) 0.5

Explanation:
1. n(N2) = 7/28 = 0.25 mol; n(O2) = 8/32 = 0.25 mol.
2. x(O2) = 0.25/0.5 = 0.5.

Teacher's Note:
1. Use molar masses 28 and 32, not mass fractions.

Question 13: In a solution of 7.8 g benzene (C6H6) and 46.0 g toluene (C6H5CH3), the mole fraction of benzene is
(a) \( \dfrac16 \)
(b) \( \dfrac15 \)
(c) \( \dfrac12 \)
(d) \( \dfrac13 \)
Show Answer & Explanation

Answer: (a) \( \dfrac16 \)

Explanation:
1. n(benzene) = 7.8/78 = 0.1; n(toluene) = 46/92 = 0.5.
2. x = 0.1/0.6 = 1/6.

Teacher's Note:
1. Molar masses: C6H6 = 78, C7H8 = 92.

Question 14: An X molal solution of a compound in water has a mole fraction of solute equal to 0.2. The value of X is
(a) 14
(b) 3.2
(c) 1.4
(d) 2
Show Answer & Explanation

Answer: (a) 14

Explanation:
1. Take 1 mole of solution: 0.2 mol solute and 0.8 mol water = 14.4 g.
2. Molality = 0.2/(14.4 × 10−3) = 13.9, about 14.

Teacher's Note:
1. Convert the mole fraction to moles per kg of solvent.

Question 15: The mole fraction of ethanol in an ethanol-water mixture is 0.25. The percentage concentration of ethanol by weight of the mixture is
(a) 25%
(b) 75%
(c) 46%
(d) 54%
Show Answer & Explanation

Answer: (c) 46%

Explanation:
1. Per mole of mixture: 0.25 mol ethanol = 11.5 g; 0.75 mol water = 13.5 g.
2. Weight % = 11.5/25 × 100 = 46%.

Teacher's Note:
1. Mole fraction and mass fraction differ because molar masses differ.

Question 16: A 500 g tooth paste sample has 0.02 g fluoride. The concentration of fluorine in ppm is
(a) 250
(b) 40
(c) 400
(d) 1000
Show Answer & Explanation

Answer: (b) 40

Explanation:
1. ppm = (0.02/500) × 106 = 40.

Teacher's Note:
1. ppm = mass of solute per million parts of sample.

Question 17: H2O2 solution used for hair bleaching is sold as approximately 5.0 g H2O2 per 100 mL of solution. The molar mass of H2O2 is 34. The molarity of this solution is approximately
(a) 0.15 M
(b) 1.5 M
(c) 3.0 M
(d) 3.4 M
Show Answer & Explanation

Answer: (b) 1.5 M

Explanation:
1. 5.0 g per 100 mL = 50 g per litre.
2. Molarity = 50/34 = 1.47, about 1.5 M.

Teacher's Note:
1. Convert per 100 mL to per litre first.

Question 18: The normality of a 10% (w/V) H2SO4 solution is nearly
(a) 0.1
(b) 0.2
(c) 0.5
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. 10% w/V = 100 g per litre = 100/98 = 1.02 M.
2. n-factor 2, so N = 2.04, about 2.

Teacher's Note:
1. Normality = molarity × 2 for H2SO4.

Question 19: What volume of 0.1 N HNO3 solution can be prepared from 6.3 g of HNO3?
(a) 1 litre
(b) 2 litre
(c) 0.5 litre
(d) 5 litre
Show Answer & Explanation

Answer: (a) 1 litre

Explanation:
1. Moles of HNO3 = 6.3/63 = 0.1 mol (n-factor 1, so 0.1 eq).
2. Volume = 0.1 mol / 0.1 M = 1 L.

Teacher's Note:
1. For monobasic acids N = M.

Question 20: Which of the following statements is true?
(a) Molarity is the number of moles of solute dissolved per litre of solvent
(b) The molarity and normality of a sodium carbonate solution are the same
(c) Molality is defined as the number of moles of solute dissolved in 1000 g of solution
(d) The ratio of the mole fraction of solute and solvent is in the ratio of their respective moles
Show Answer & Explanation

Answer: (d) The ratio of the mole fraction of solute and solvent is in the ratio of their respective moles

Explanation:
1. xsolute : xsolvent = n : N because both have the same denominator.
2. Statements (a) and (c) wrongly use litre of solvent and gram of solution, and for Na2CO3 N = 2M.

Teacher's Note:
1. Molarity uses solution volume; molality uses solvent mass.

Question 21: The molarity of 98% H2SO4 (d = 1.8 g mL−1) by weight is
(a) 6 M
(b) 18 M
(c) 10 M
(d) 4 M
Show Answer & Explanation

Answer: (b) 18 M

Explanation:
1. 1000 mL has mass 1800 g, containing 0.98 × 1800 = 1764 g H2SO4.
2. Moles = 1764/98 = 18, so molarity = 18 M.

Teacher's Note:
1. M = (10 × d × wt%)/molar mass.

Question 22: 6.02 × 1020 molecules of urea are present in 100 mL of a solution. The concentration of the solution is
(a) 0.1 M
(b) 0.02 M
(c) 0.01 M
(d) 0.001 M
Show Answer & Explanation

Answer: (c) 0.01 M

Explanation:
1. Moles = 6.02 × 1020/6.02 × 1023 = 10−3 mol.
2. Molarity = 10−3/0.1 L = 0.01 M.

Teacher's Note:
1. Use Avogadro's number to convert molecules to moles.

Question 23: If the density of methanol is 0.8 kg L−1, the volume of methanol needed to make 2.5 L of its 0.4 M solution is
(a) 0.4 L
(b) 4.0 L
(c) 0.04 L
(d) 40 L
Show Answer & Explanation

Answer: (c) 0.04 L

Explanation:
1. Moles needed = 0.4 × 2.5 = 1.0 mol.
2. Mass = 32 g, volume = 32/800 g L−1 = 0.04 L.

Teacher's Note:
1. 0.8 kg L−1 = 800 g L−1.

Question 24: The molality of a solution containing 3 g of CH3COOH in 50 g of benzene is
(a) 0.1 m
(b) 0.5 m
(c) 0.2 m
(d) 1 m
Show Answer & Explanation

Answer: (d) 1 m

Explanation:
1. Moles = 3/60 = 0.05.
2. Molality = 0.05/0.050 kg = 1 m.

Teacher's Note:
1. Mass of solvent must be in kg.

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