Question 41 of 50advanced🔧 ApplyNumerical3 marks

Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4×1.4 \times 10^{-3}$$, Kf = 1.86 K kg mol-1.

Correct Answer

The depression in freezing point is 0.65 K.

Exercise: EXERCISES | Q: 1.32 | (Page 29)
For More Understanding

Explanation

This problem involves calculating freezing point depression for a weak acid solution. Since CH₃CH₂CHClCOOH is a weak acid, it partially dissociates in water, requiring consideration of the van't Hoff factor. The degree of dissociation (α) is calculated using the dissociation constant Ka, and then applied to find the effective molality through the van't Hoff factor i = 1 + α.

Solution Steps

  1. Step 1: Calculate molar mass of CH₃CH₂CHClCOOH: M = (4×124 \times 12) + (7×17 \times 1) + 35.5 + (2×162 \times 16) = 48 + 7 + 35.5 + 32 = 122.5 g mol⁻¹

  2. Step 2: Calculate molality of solution: Moles of solute = 10 g / 122.5 g mol⁻¹ = 0.0816 mol. Mass of solvent = 250 g = 0.25 kg. Molality (m) = 0.0816 mol / 0.25 kg = 0.326 mol kg⁻¹

  3. Step 3: Calculate degree of dissociation (α): For weak acid, Ka = cα²/(1-α) ≈\approx cα² (for small α). α = Ka/c\sqrt{Ka/c} = \sqrt{$$1.4 \times 10$$⁻³ / 0.326} = 0.0043\sqrt{0.0043} = 0.065

  4. Step 4: Calculate van't Hoff factor (i): For monoprotic acid dissociating into 2 ions, i = 1 + α = 1 + 0.065 = 1.065

  5. Step 5: Calculate depression in freezing point: ΔTf = i × Kf × m = 1.065×1.861.065 \times 1.86 × 0.326 = 0.65 K

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