JEE Chemistry Solutions MCQs Set 14

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Multiple Choice Questions

Question 1: Equal volumes of 0.1 M urea and 0.1 M glucose are mixed. The mixture will have:
(a) the same osmotic pressure
(b) none of these
(c) higher osmotic pressure
(d) lower osmotic pressure
Show Answer & Explanation

Answer: (a) the same osmotic pressure

Explanation:
1. Mixing equal volumes halves each concentration to 0.05 M, so the total particle concentration is 0.1 M, the same as before.
2. The osmotic pressure is therefore the same as that of each original solution.

Teacher's Note:
1. For non-electrolytes, osmotic pressure depends on the total molar concentration of particles.

Question 2: A 0.0020 m aqueous solution of an ionic compound [Co(NH3)5(NO2)]Cl freezes at -0.00732 °C. The number of moles of ions which 1 mol of the ionic compound produces on being dissolved in water will be (Kf = 1.86 °C m⁻¹):
(a) 1
(b) 2
(c) 4
(d) 3
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. \( i = \dfrac{\Delta T_f}{K_fm} = \dfrac{0.00732}{1.86\times0.0020} = 1.97\approx2 \).
2. The compound gives \( [Co(NH_3)_5(NO_2)]^+ \) and \( Cl^- \), which is 2 ions.

Teacher's Note:
1. The van't Hoff factor of a strong electrolyte is the number of ions per formula unit.

Question 3: The molality of a urea solution in which 0.0100 g of urea [(NH2)2CO] is added to 0.3000 dm³ of water at STP is:
(a) 5.55 × 10⁻⁴ m
(b) 0.555 m
(c) 3.33 × 10⁻² m
(d) 33.3 m
Show Answer & Explanation

Answer: (a) 5.55 × 10⁻⁴ m

Explanation:
1. Moles of urea \( = \dfrac{0.0100}{60} = 1.67\times10^{-4} \).
2. The mass of 0.3 dm³ of water is 0.3 kg, so \( m = \dfrac{1.67\times10^{-4}}{0.3} = 5.55\times10^{-4} \) m.

Teacher's Note:
1. 1 dm³ = 1 L, and water has a density of 1 kg/L.

Question 4: The molarity of a 0.2 N Na2CO3 solution will be:
(a) 0.05 M
(b) 0.1 M
(c) 0.2 M
(d) 0.4 M
Show Answer & Explanation

Answer: (b) 0.1 M

Explanation:
1. For Na2CO3 the n-factor is 2.
2. \( M = \dfrac{N}{n} = \dfrac{0.2}{2} = 0.1 \) M.

Teacher's Note:
1. Divide normality by the n-factor to get molarity.

Question 5: The osmotic pressure at 15 °C of a solution (density 1 g mL⁻¹) containing 3 g of glucose (molecular weight = 180) in 60 g of water is:
(a) 6.25 atm
(b) 0.34 atm
(c) 0.65 atm
(d) 5.57 atm
Show Answer & Explanation

Answer: (a) 6.25 atm

Explanation:
1. The mass of solution is 63 g, so its volume is 63 mL = 0.063 L.
2. \( C = \dfrac{3/180}{0.063} = 0.2646 \) M.
3. \( \pi = CRT = 0.2646\times0.0821\times288 \approx 6.25 \) atm.

Teacher's Note:
1. The volume of the solution (not the solvent) is used for the molarity.

Question 6: An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?
(a) addition of 1.00 molal KI
(b) addition of NaCl
(c) addition of water
(d) addition of Na2SO4
Show Answer & Explanation

Answer: (c) addition of water

Explanation:
1. Adding water dilutes the solution, so the mole fraction of the solvent increases.
2. The vapour pressure of the solution therefore rises towards that of pure water.

Teacher's Note:
1. Adding more solute lowers the vapour pressure. Adding solvent raises it.

Question 7: The concentration of a solution of H2O2 is 6.8% (w/V). The volume strength of the solution is:
(a) 22.4
(b) 5
(c) 11.2
(d) 20
Show Answer & Explanation

Answer: (a) 22.4

Explanation:
1. 6.8% w/V means 68 g per litre, so \( N = \dfrac{68}{17} = 4 \).
2. Volume strength \( = 5.6\times4 = 22.4 \).

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

Question 8: The normality of a 0.3 M phosphorous acid (H3PO3) solution is:
(a) 0.9
(b) 0.1
(c) 0.15
(d) 0.6
Show Answer & Explanation

Answer: (d) 0.6

Explanation:
1. H3PO3 is dibasic (only two of the three H atoms are acidic), so n = 2.
2. \( N = 0.3\times2 = 0.6 \).

Teacher's Note:
1. Phosphorous acid is H3PO3 with two replaceable hydrogens, unlike phosphoric acid H3PO4.

Question 9: The osmotic pressure of a sugar solution at 24 °C is 2.5 atmosphere. The concentration of the solution in mole per litre is:
(a) 0.1025
(b) 1025
(c) 10.25
(d) 1.025
Show Answer & Explanation

Answer: (a) 0.1025

Explanation:
1. \( C = \dfrac{\pi}{RT} = \dfrac{2.5}{0.0821\times297} = 0.1025 \) mol/L.

Teacher's Note:
1. Convert the temperature to kelvin (24 + 273 = 297 K).

Question 10: A solution of sucrose (molar mass = 342 g mol⁻¹) is made by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol⁻¹):
(a) +0.372 °C
(b) -0.372 °C
(c) -0.520 °C
(d) -0.570 °C
Show Answer & Explanation

Answer: (b) -0.372 °C

Explanation:
1. \( m = \dfrac{68.5}{342} = 0.2 \).
2. \( \Delta T_f = 1.86\times0.2 = 0.372 \), so the freezing point is \( -0.372 \) °C.

Teacher's Note:
1. Sucrose is a non-electrolyte, so i = 1.

Question 11: A gaseous mixture is made by taking equal moles of CO and N2. If the total pressure of the mixture was found to be 1 atmosphere, the partial pressure of nitrogen (N2) in the mixture is:
(a) 0.9 atm
(b) 0.8 atm
(c) 1 atm
(d) 0.5 atm
Show Answer & Explanation

Answer: (d) 0.5 atm

Explanation:
1. The mole fraction of N2 is 0.5.
2. \( p_{N_2} = 0.5\times1 = 0.5 \) atm.

Teacher's Note:
1. Partial pressure = mole fraction × total pressure.

Question 12: The molarity of pure water is:
(a) 100 M
(b) 50 M
(c) 18 M
(d) 55.5 M
Show Answer & Explanation

Answer: (d) 55.5 M

Explanation:
1. Take 1 L of water: mass 1000 g, moles \( = 1000/18 = 55.5 \).
2. Molarity = 55.5 M.

Teacher's Note:
1. Water has a density of 1 g/mL and a molar mass of 18 g/mol.

Question 13: A solution containing 4 g of a non-volatile organic solute per 100 mL was found to have an osmotic pressure equal to 500 cm of mercury at 27 °C. The molecular weight of the solute is:
(a) 149.7
(b) 14.97
(c) 1697
(d) 1.497
Show Answer & Explanation

Answer: (a) 149.7

Explanation:
1. \( \pi = 500\ \text{cm Hg} = \dfrac{500}{76} = 6.58 \) atm.
2. \( C = \dfrac{6.58}{0.0821\times300} = 0.267 \) M. The solution contains 40 g per litre.
3. \( M = \dfrac{40}{0.267} \approx 149.7 \) g/mol.

Teacher's Note:
1. 76 cm of mercury is equal to 1 atm, so convert the pressure to atmospheres before using R = 0.0821.

Question 14: For an ideal solution of A and B, which statement is incorrect?
(a) The entropy change of the mixture of A and B is zero
(b) The volume change of the solution of A and B is zero
(c) The intermolecular forces of A and B are the same as those of A-A and B-B
(d) The enthalpy change of mixing of A and B is zero
Show Answer & Explanation

Answer: (a) The entropy change of the mixture of A and B is zero

Explanation:
1. For an ideal solution ΔH(mix) = 0 and ΔV(mix) = 0, and the A-B forces equal the A-A and B-B forces.
2. The entropy of mixing is positive, so statement (4) is incorrect.

Teacher's Note:
1. Spontaneous mixing of ideal liquids is driven entirely by an increase in entropy.

Question 15: The vapour pressures of two liquids 'P' and 'Q' are 80 and 60 torr respectively. The total vapour pressure of the solution obtained by mixing 3 mol of P and 2 mol of Q would be:
(a) 72 torr
(b) 20 torr
(c) 68 torr
(d) 140 torr
Show Answer & Explanation

Answer: (a) 72 torr

Explanation:
1. \( x_P = 0.6 \), \( x_Q = 0.4 \).
2. \( P = 0.6\times80+0.4\times60 = 48+24 = 72 \) torr.

Teacher's Note:
1. The answer must lie between 60 and 80 torr.

Question 16: 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 more than that of KCl
(c) that of NaCl will be half of that of KCl
(d) that of NaCl will be less than that of KCl
Show Answer & Explanation

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

Explanation:
1. Moles = mass / molar mass. NaCl (58.5) has a smaller molar mass than KCl (74.5), so equal masses contain more moles of NaCl.
2. Same volume, so the molarity of NaCl is greater.

Teacher's Note:
1. Equal masses mean more moles of the substance with the lower molar mass.

Question 17: A solution containing 8.6 g of urea in one litre was found to be isotonic with a 0.5% (w/V) solution of an organic non-volatile solute. The molecular weight of the latter is:
(a) 348.9
(b) 3489
(c) 861.2
(d) 34.89
Show Answer & Explanation

Answer: (d) 34.89

Explanation:
1. Isotonic solutions have equal molarity: \( \dfrac{8.6}{60\times1} = \dfrac{5}{M\times1} \) (0.5% means 5 g per litre).
2. \( M = \dfrac{5\times60}{8.6} = 34.9 \) g/mol.

Teacher's Note:
1. Isotonic means equal osmotic pressure at the same temperature, so equal molar concentration.

Question 18: The freezing point depression constant for water is 1.86 °C m⁻¹. If 5.0 g of Na2SO4 is dissolved in 45.0 g of H2O, the freezing point is changed by 3.82 °C. Calculate the van't Hoff factor for Na2SO4:
(a) 2.63
(b) 0.381
(c) 2.05
(d) 3.11
Show Answer & Explanation

Answer: (a) 2.63

Explanation:
1. \( m = \dfrac{5.0/142}{0.045} = 0.7825 \) mol/kg.
2. \( i = \dfrac{\Delta T_f}{K_fm} = \dfrac{3.82}{1.86\times0.7825} = 2.62 \approx 2.63 \).

Teacher's Note:
1. Na2SO4 can give at most i = 3, so the answer must be below 3.

Question 19: Which of the following plots does NOT represent the behaviour of an ideal binary liquid solution?
(a) Plot of Ptotal versus XA is non-linear
(b) Plot of Ptotal versus XA (or XB) is linear
(c) Plot of PB versus XB is linear
(d) Plot of PA versus XA (mole fraction of A in liquid phase) is linear
Show Answer & Explanation

Answer: (a) Plot of Ptotal versus XA is non-linear

Explanation:
1. For an ideal solution \( P_A = X_AP_A^0 \) and \( P_B = X_BP_B^0 \) are linear.
2. \( P_{total} = P_B^0+X_A(P_A^0-P_B^0) \) is also linear in \( X_A \), so the statement that it is non-linear is incorrect.

Teacher's Note:
1. For a non-ideal solution, the total pressure curve bends upwards or downwards.

Question 20: The mole fraction of oxygen in a mixture of 7 g of nitrogen and 8 g of oxygen is:
(a) 0.5
(b) 0.25
(c) 1.0
(d) \( \dfrac{8}{15} \)
Show Answer & Explanation

Answer: (a) 0.5

Explanation:
1. Moles of N2 \( = 7/28 = 0.25 \). Moles of O2 \( = 8/32 = 0.25 \).
2. \( x_{O_2} = \dfrac{0.25}{0.5} = 0.5 \).

Teacher's Note:
1. Use molar masses 28 for N2 and 32 for O2, not the atomic masses.

Question 21: If a 6.84% (w/V) solution of cane sugar (mol. wt. 342) is isotonic with a 1.52% (w/V) solution of thiocarbamide, then the molecular weight of thiocarbamide is:
(a) 76
(b) 60
(c) 152
(d) 180
Show Answer & Explanation

Answer: (a) 76

Explanation:
1. Sugar: 68.4 g per litre, \( C = 68.4/342 = 0.2 \) M.
2. Thiocarbamide: 15.2 g per litre, so \( \dfrac{15.2}{M} = 0.2 \Rightarrow M = 76 \).

Teacher's Note:
1. Equal osmotic pressure means equal molar concentration for non-electrolytes.

Question 22: The van't Hoff factor i for a compound which undergoes dissociation in one solvent and association in another solvent is respectively:
(a) greater than one and less than one
(b) less than one and greater than one
(c) less than one and less than one
(d) greater than one and greater than one
Show Answer & Explanation

Answer: (a) greater than one and less than one

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

Teacher's Note:
1. Remember: dissociation gives i greater than 1, association gives i less than 1.

Question 23: The vapour pressures of two pure liquids (A) and (B) are 100 and 80 torr respectively. The total pressure of the solution obtained by mixing 2 mol of (A) and 3 mol of (B) would be:
(a) 20 torr
(b) 88 torr
(c) 36 torr
(d) 180 torr
Show Answer & Explanation

Answer: (b) 88 torr

Explanation:
1. \( x_A = 0.4 \), \( x_B = 0.6 \).
2. \( P = 0.4\times100+0.6\times80 = 40+48 = 88 \) torr.

Teacher's Note:
1. The total pressure lies between the two pure vapour pressures.

Question 24: In a solution of 7.8 g of benzene (C6H6) and 46.0 g of toluene (C6H5CH3), the mole fraction of benzene is:
(a) \( \dfrac16 \)
(b) \( \dfrac12 \)
(c) \( \dfrac15 \)
(d) \( \dfrac13 \)
Show Answer & Explanation

Answer: (a) \( \dfrac16 \)

Explanation:
1. Benzene: \( 7.8/78 = 0.1 \) mol. Toluene: \( 46/92 = 0.5 \) mol.
2. \( x_{benzene} = \dfrac{0.1}{0.6} = \dfrac16 \).

Teacher's Note:
1. The molar masses are 78 for benzene and 92 for toluene.

Question 25: A solution containing 500 g of a protein per litre is isotonic with a solution containing 3.42 g of sucrose per litre. The molecular mass of the protein is:
(a) 146
(b) 5
(c) 34200
(d) 50000
Show Answer & Explanation

Answer: (d) 50000

Explanation:
1. Sucrose: \( 3.42/342 = 0.01 \) M.
2. Protein: \( \dfrac{500}{M} = 0.01 \Rightarrow M = 50000 \).

Teacher's Note:
1. Equal molarity of the protein and sucrose solutions gives the molar mass directly.

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