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Solutions question

2007 · Shift 1 · Q8
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Solutions question

2007 · Shift 1 · Q8

JEE AdvancedChemistrySolutionsMCQ+3 / −1
When 20 g of naphthoic acid (C 11{}_{11}11​ H 8{}_{8}8​ O 2{}_{2}2​) is dissolved in 50 g of benzene in 50 g of benzene (K f{}_ff​ = 1.72 K kg mol −1{}^{-1}−1), a freezing point depression of 2 K is observed. The van't Hoff factor (i) is :
  1. A
    0.5
  2. B
    1
  3. C
    2
  4. D
    3
View written solutionFree

Correct answer: A

Step-by-step Solution:

  1. Identify the given data and the required quantity.

    • Mass of solute (naphthoic acid, C₁₁H₈O₂), w₂ = 20 g.
    • Mass of solvent (benzene), w₁ = 50 g = 0.050 kg.
    • Freezing point depression, ΔTf=2ΔT_f = 2ΔTf​=2 K.
    • Molal freezing point depression constant for benzene, Kf=1.72K_f = 1.72Kf​=1.72 K kg mol⁻¹.
    • We need to find the van't Hoff factor, i.
  2. Calculate the molar mass of the solute (naphthoic acid, C₁₁H₈O₂).

    • The chemical formula is C₁₁H₈O₂.
    • Molar mass M₂ = (11 × ext{Atomic mass of C}) + (8 × ext{Atomic mass of H}) + (2 × ext{Atomic mass of O})
    • M₂ = (11 × 12.01) + (8 × 1.008) + (2 × 16.00)
    • M₂ = 132.11 + 8.064 + 32.00 = 172.174 g/mol. For simplicity, we can use integer masses: M₂ = 11×12 + 8×1 + 2×16 = 132 + 8 + 32 = 172 g/mol.
  3. Calculate the theoretical molality (m) of the solution.

    • Molality is defined as the number of moles of solute per kilogram of solvent.
    • m = (moles of solute) / (mass of solvent in kg)
    • Moles of solute n₂ = w₂ / M₂ = 20 g / 172 g/mol ≈ 0.1163 mol.
    • Mass of solvent w₁ = 50 g = 0.050 kg.
    • m = 0.1163 mol / 0.050 kg ≈ 2.326 mol/kg.
  4. Use the formula for freezing point depression including the van't Hoff factor.

    • The formula for the depression in freezing point is: ΔTf=i×Kf×mΔT_f = i × K_f × mΔTf​=i×Kf​×m
    • We need to solve for i: i=ΔTfKf×mi = \frac{ΔT_f}{K_f × m}i=Kf​×mΔTf​​
  5. Substitute the known values into the formula to find 'i'.

    • ΔTf=2ΔT_f = 2ΔTf​=2 K
    • Kf=1.72K_f = 1.72Kf​=1.72 K kg mol⁻¹
    • m = 2.326 mol/kg
    • i=21.72×2.326i = \frac{2}{1.72 × 2.326}i=1.72×2.3262​
    • i=24.00072i = \frac{2}{4.00072}i=4.000722​
    • i≈0.4999≈0.5i ≈ 0.4999 ≈ 0.5i≈0.4999≈0.5
  6. Interpret the result and select the correct option.

    • The calculated van't Hoff factor i is approximately 0.5.
    • A van't Hoff factor less than 1 indicates that the solute particles are associating in the solvent. In this case, since i = 0.5, it suggests that naphthoic acid (a carboxylic acid) forms dimers in the non-polar solvent benzene, likely through hydrogen bonding. For complete dimerization, i would be exactly 0.5.
    • Comparing our result with the options:
      • A: 0.5
      • B: 1
      • C: 2
      • D: 3
    • The correct option is A.
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