MCQ
The solution from the following with highest depression in freezing point/lowest freezing point is
  • $180 \mathrm{~g}$ of acetic acid dissolved in water
  • B
    $180 \mathrm{~g}$ of acetic acid dissolved in benzene
  • C
    $180 \mathrm{~g}$ of benzoic acid dissolved in benzene
  • D
    $180 \mathrm{~g}$ of glucose dissolved in water

Answer

Correct option: A.
$180 \mathrm{~g}$ of acetic acid dissolved in water
a
$\Delta \mathrm{T}_{\mathrm{f}}$ is maximum when $\mathrm{i} \times \mathrm{m}$ is maximum.

1) $\mathrm{m}_1=\frac{180}{60}=3, \mathrm{i}=1+\alpha$

Hence

$\Delta \mathrm{T}_{\mathrm{f}}=(1+\alpha) \cdot \mathrm{k}_{\mathrm{f}}=3 \times 1.86=5.58^{\circ} \mathrm{C}(\alpha<<1)$

$2)$ $\mathrm{m}_2=\frac{180}{60}=3, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{3}{2} \times \mathrm{k}_{\mathrm{f}}{ }^{\prime}=7.68^{\circ} \mathrm{C}$

3) $\mathrm{m}_3=\frac{180}{122}=1.48, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{1.48}{2} \times \mathrm{k}_{\mathrm{f}}^{\prime}=3.8^{\circ} \mathrm{C}$

4) $\mathrm{m}_4=\frac{180}{180}=1, \mathrm{i}=1, \Delta \mathrm{T}_{\mathrm{f}}=1 \cdot \mathrm{k}_{\mathrm{f}}^{\prime}=1.86^{\circ} \mathrm{C}$

As per NCERT, $\mathrm{k}_{\mathrm{f}}{ }^{\prime}\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{k} \cdot \mathrm{kg} \mathrm{mol}^{-1}$

$\mathrm{k}_{\mathrm{f}}{ }^{\prime}($ Benzene $)=5.12 \mathrm{k} \cdot \mathrm{kg} \mathrm{mol}^{-1}$

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