JEE Chemistry Solutions MCQs Set 15

Multiple Choice Questions (MCQs) for JEE Chemistry: Solutions

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Practice Solutions MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: A 500 g tooth paste sample has 0.02 g fluoride. What is the concentration of fluorine in terms of ppm?
(a) 1000
(b) 400
(c) 250
(d) 40
Show Answer & Explanation

Answer: (d) 40

Explanation:
1. \( \text{ppm} = \dfrac{0.02}{500}\times10^6 = 40 \).

Teacher's Note:
1. ppm = (mass of component / total mass) × 10⁶.

Question 2: The van't Hoff factor is:
(a) less than one in case of association
(b) always less than one
(c) less than one in case of dissociation
(d) more than one in case of association
Show Answer & Explanation

Answer: (a) less than one in case of association

Explanation:
1. Association decreases the number of particles, so i < 1.
2. Dissociation increases the number of particles, so i > 1.

Teacher's Note:
1. i = observed colligative property / calculated colligative property.

Question 3: Which of the following compounds is best suited for use as an antifreeze in automobile radiators?
(a) Ethyl alcohol
(b) Glycol
(c) Methyl alcohol
(d) Nitrophenol
Show Answer & Explanation

Answer: (b) Glycol

Explanation:
1. An antifreeze should dissolve in water, lower its freezing point and have a high boiling point so that it does not evaporate.
2. Ethylene glycol is non-volatile at radiator temperatures, whereas the simple alcohols evaporate easily.

Teacher's Note:
1. Ethylene glycol is the usual antifreeze because of its low volatility and complete miscibility.

Question 4: The vapour pressures of ethanol and methanol are 42.0 mm and 88.5 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 46.0 g of ethanol with 16.0 g of methanol. The mole fraction of methanol in the vapour is:
(a) 0.502
(b) 0.556
(c) 0.467
(d) 0.513
Show Answer & Explanation

Answer: (d) 0.513

Explanation:
1. Ethanol: \( 46/46 = 1 \) mol. Methanol: \( 16/32 = 0.5 \) mol, so \( x_M = 1/3 \), \( x_E = 2/3 \).
2. \( p_M = 29.5 \), \( p_E = 28 \), total 57.5.
3. \( y_M = 29.5/57.5 = 0.513 \).

Teacher's Note:
1. Convert masses to moles, then use Raoult's law for the partial pressures.

Question 5: The normality of a 10 mL of '20 V' H2O2 solution is:
(a) 1.79
(b) 6.086
(c) 3.58
(d) 60.86
Show Answer & Explanation

Answer: (c) 3.58

Explanation:
1. The strength in volume 20 V means 1 mL gives 20 mL of O2 at STP, so \( N = \dfrac{20}{5.6} \).
2. \( N = 3.57 \approx 3.58 \).

Teacher's Note:
1. Normality = volume strength / 5.6. The volume of the sample taken does not change the normality.

Question 6: The van't Hoff factor (i) for a dilute solution of K3[Fe(CN)6] is:
(a) 10
(b) 0.25
(c) 4
(d) 5
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. \( K_3[Fe(CN)_6] \to 3K^+ + [Fe(CN)_6]^{3-} \), which gives 4 ions.
2. For complete dissociation, i = 4.

Teacher's Note:
1. The complex ion [Fe(CN)6]3− does not break up further.

Question 7: The vapour pressures of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 25 °C are 200 mm Hg and 415 mm Hg respectively. The vapour pressure of the solution obtained by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at the same temperature will be (molecular mass of CHCl3 = 119.5 u and of CH2Cl2 = 85 u):
(a) 280.5 mm Hg
(b) 615 mm Hg
(c) 347.9 mm Hg
(d) 173.9 mm Hg
Show Answer & Explanation

Answer: (c) 347.9 mm Hg

Explanation:
1. Moles: \( \text{CHCl}_3 = 25.5/119.5 = 0.2134 \), \( \text{CH}_2\text{Cl}_2 = 40/85 = 0.4706 \). Total \( = 0.684 \).
2. \( x_1 = 0.312 \), \( x_2 = 0.688 \).
3. \( P = 0.312\times200+0.688\times415 = 62.4+285.5 = 347.9 \) mm Hg.

Teacher's Note:
1. The answer must lie between the two pure vapour pressures, 200 and 415 mm Hg.

Question 8: A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20 °C are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be:
(a) 0.786
(b) 0.200
(c) 0.478
(d) 0.549
Show Answer & Explanation

Answer: (c) 0.478

Explanation:
1. \( x_{pentane} = 0.2 \). \( p_{pentane} = 88 \), \( p_{hexane} = 0.8\times120 = 96 \), total 184 mm Hg.
2. \( y = \dfrac{88}{184} = 0.478 \).

Teacher's Note:
1. The vapour is enriched in the more volatile pentane, so 0.478 is greater than 0.2.

Question 9: H2O2 solution used for hair bleaching is sold as a solution of approximately 5.0 g H2O2 per 100 mL of the solution. The molecular mass of H2O2 is 34. The molarity of this solution is approximately:
(a) 0.15 M
(b) 1.5 M
(c) 3.4 M
(d) 3.0 M
Show Answer & Explanation

Answer: (b) 1.5 M

Explanation:
1. 5.0 g in 100 mL means 50 g per litre.
2. \( M = \dfrac{50}{34} \approx 1.5 \) M.

Teacher's Note:
1. Scale the mass to one litre of solution and divide by the molar mass.

Question 10: The experimental molecular weight of an electrolyte will always be less than its calculated value because the value of the van't Hoff factor i is:
(a) less than 1
(b) zero
(c) one
(d) greater than 1
Show Answer & Explanation

Answer: (d) greater than 1

Explanation:
1. \( M_{observed} = \dfrac{M_{calculated}}{i} \).
2. For an electrolyte \( i > 1 \), so the observed molar mass is smaller.

Teacher's Note:
1. Dissociation makes the apparent molar mass smaller.

Question 11: \( 6.02\times10^{20} \) molecules of urea are present in 100 mL of its solution. The concentration of the solution is:
(a) 0.01 M
(b) 0.1 M
(c) 0.02 M
(d) 0.001 M
Show Answer & Explanation

Answer: (a) 0.01 M

Explanation:
1. Moles \( = \dfrac{6.02\times10^{20}}{6.02\times10^{23}} = 10^{-3} \).
2. \( M = \dfrac{10^{-3}}{0.1\ \text{L}} = 0.01 \) M.

Teacher's Note:
1. Divide the number of molecules by Avogadro's number to get moles.

Question 12: A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be:
(a) 300
(b) 700
(c) 360
(d) 350
Show Answer & Explanation

Answer: (d) 350

Explanation:
1. \( 290 = 0.6P_E^0 + 0.4\times200 \).
2. \( 0.6P_E^0 = 210 \Rightarrow P_E^0 = 350 \) mm.

Teacher's Note:
1. The pure ethanol vapour pressure must be higher than the mixture's 290 mm because propanol is less volatile.

Question 13: 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 means 10 g in 100 mL = 100 g per litre.
2. Equivalent mass of H2SO4 \( = 49 \), so \( N = \dfrac{100}{49} \approx 2 \).

Teacher's Note:
1. Normality = (grams per litre) / equivalent mass.

Question 14: The van't Hoff factor (i) for a dilute aqueous solution of glucose is:
(a) 1.5
(b) zero
(c) 1.0
(d) 2.0
Show Answer & Explanation

Answer: (c) 1.0

Explanation:
1. Glucose is a non-electrolyte and neither dissociates nor associates in water.
2. So i = 1.0.

Teacher's Note:
1. Non-electrolytes have i = 1.

Question 15: The boiling point of benzene is 353.23 K. When 1.8 g of a non-volatile solute is dissolved in 90 g of benzene, the boiling point rises to 354.11 K. Given Kb (benzene) = 2.53 K kg mol⁻¹, the molecular mass of the non-volatile substance is:
(a) 120 g mol⁻¹
(b) 116 g mol⁻¹
(c) 58 g mol⁻¹
(d) 60 g mol⁻¹
Show Answer & Explanation

Answer: (c) 58 g mol⁻¹

Explanation:
1. \( \Delta T_b = 354.11-353.23 = 0.88 \) K.
2. \( M = \dfrac{K_b\times w\times1000}{\Delta T_b\times W} = \dfrac{2.53\times1.8\times1000}{0.88\times90} = 57.5\approx58 \) g/mol.

Teacher's Note:
1. Use the molar mass formula with the solvent mass in grams and the factor 1000.

Question 16: A mixture of liquids showing positive deviation from Raoult's law is:
(a) (CH3)2CO + C6H5NH2
(b) (CH3)2CO + C2H5OH
(c) (C2H5)2O + CHCl3
(d) (CH3)2CO + CHCl3
Show Answer & Explanation

Answer: (b) (CH3)2CO + C2H5OH

Explanation:
1. Acetone with ethanol: ethanol's hydrogen bonds are broken, so the A-B attraction is weaker than A-A and B-B, giving a positive deviation.
2. The other three pairs form hydrogen bonds between the two components and show a negative deviation.

Teacher's Note:
1. Positive deviation arises when the new interactions are weaker than the old ones.

Question 17: What volume of a 0.1 N HNO3 solution can be prepared from 6.3 g of HNO3?
(a) 5 litre
(b) 0.5 litre
(c) 2 litre
(d) 1 litre
Show Answer & Explanation

Answer: (d) 1 litre

Explanation:
1. Equivalent mass of HNO3 \( = 63 \), so 6.3 g \( = 0.1 \) equivalent.
2. \( V = \dfrac{0.1\ \text{eq}}{0.1\ \text{N}} = 1 \) litre.

Teacher's Note:
1. Equivalents = mass / equivalent mass. Volume = equivalents / normality.

Question 18: A 0.585% NaCl solution at 27 °C has an osmotic pressure of (assume complete dissociation):
(a) 3.8 atm
(b) 1.2 atm
(c) 2.49 atm
(d) 5 atm
Show Answer & Explanation

Answer: (d) 5 atm

Explanation:
1. 0.585% means 5.85 g per litre, \( C = 5.85/58.5 = 0.1 \) M.
2. \( \pi = iCRT = 2\times0.1\times0.0821\times300 = 4.93 \approx 5 \) atm.

Teacher's Note:
1. NaCl gives two ions, so i = 2.

Question 19: The strength of H2O2 is 15.18 g L⁻¹. It is equal to:
(a) 10 volume
(b) 5 volume
(c) 1 volume
(d) 7 volume
Show Answer & Explanation

Answer: (b) 5 volume

Explanation:
1. Normality \( = \dfrac{15.18}{17} = 0.893 \) N (equivalent mass of H2O2 = 17).
2. Volume strength \( = 5.6\times0.893 = 5.0 \). It is a '5 volume' solution.

Teacher's Note:
1. Volume strength = 5.6 × normality.

Question 20: An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressures of methanol and ethanol are 2.619 kPa and 4.556 kPa respectively, the composition of the vapour (in terms of mole fraction) will be:
(a) 0.635 MeOH, 0.365 EtOH
(b) 0.365 MeOH, 0.635 EtOH
(c) 0.574 MeOH, 0.326 EtOH
(d) 0.173 MeOH, 0.827 EtOH
Show Answer & Explanation

Answer: (b) 0.365 MeOH, 0.635 EtOH

Explanation:
1. Total \( = 2.619+4.556 = 7.175 \) kPa.
2. \( y_{MeOH} = \dfrac{2.619}{7.175} = 0.365 \) and \( y_{EtOH} = 0.635 \).

Teacher's Note:
1. The mole fraction in the vapour is the partial pressure divided by the total pressure.

Question 21: A molal solution is one that contains one mole of a solute in:
(a) one litre of the solution
(b) 1000 g of the solvent
(c) one litre of the solvent
(d) 22.4 litres of the solution
Show Answer & Explanation

Answer: (b) 1000 g of the solvent

Explanation:
1. Molality is the number of moles of solute per 1000 g (1 kg) of the solvent.

Teacher's Note:
1. Molality depends on the mass of the solvent, so it does not change with temperature.

Question 22: A substance A when dissolved in a solvent B shows the molecular mass corresponding to A3. The van't Hoff factor will be:
(a) 1
(b) \( \dfrac13 \)
(c) 3
(d) 2
Show Answer & Explanation

Answer: (b) \( \dfrac13 \)

Explanation:
1. Three molecules of A associate into one particle, so the number of particles decreases to one-third.
2. \( i = \dfrac13 \).

Teacher's Note:
1. Association into trimers makes the observed molar mass three times larger, so i = 1/3.

Question 23: Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?
(a) Al2(SO4)3
(b) C6H12O6
(c) K2SO4
(d) KCl
Show Answer & Explanation

Answer: (a) Al2(SO4)3

Explanation:
1. Number of ions: KCl 2, glucose 1, Al2(SO4)3 5 and K2SO4 3.
2. The largest van't Hoff factor, i = 5 for Al2(SO4)3, gives the largest depression.

Teacher's Note:
1. At equal molality, the depression follows the van't Hoff factor.

Question 24: Azeotropic mixtures are:
(a) constant boiling mixtures
(b) those which boil at different temperatures
(c) mixtures of two solids
(d) those which can be fractionally distilled
Show Answer & Explanation

Answer: (a) constant boiling mixtures

Explanation:
1. An azeotrope boils at a constant temperature, and the vapour has the same composition as the liquid.
2. So it cannot be separated by fractional distillation.

Teacher's Note:
1. The word azeotrope means 'boiling without change'.

Question 25: 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 solution of sodium carbonate are the same. (c) Molality (m) of a solution is defined as the number of moles of solute dissolved in 1000 g of solution. (d) The ratio of the mole fractions of solute and solvent is in the ratio of their respective moles.
(a) (a) and (c)
(b) (b) and (c)
(c) (a) and (d)
(d) only (d)
Show Answer & Explanation

Answer: (d) only (d)

Explanation:
1. (a) is false: molarity is per litre of solution. (b) is false: for Na2CO3, \( N = 2M \). (c) is false: molality is per 1000 g of solvent.
2. (d) is true: \( \dfrac{x_1}{x_2} = \dfrac{n_1/(n_1+n_2)}{n_2/(n_1+n_2)} = \dfrac{n_1}{n_2} \).

Teacher's Note:
1. The common denominator (total moles) cancels in a ratio of mole fractions.

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