Solutions Question 11
Question 11 - 2024 (30 Jan Shift 2)
The solution from the following with highest depression in freezing point/lowest freezing point is
(1) $180 \mathrm{~g}$ of acetic acid dissolved in water
(2) $180 \mathrm{~g}$ of acetic acid dissolved in benzene
(3) $180 \mathrm{~g}$ of benzoic acid dissolved in benzene
(4) $180 \mathrm{~g}$ of glucose dissolved in water
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Answer (1)
Solution
$\Delta \mathrm{T}_{\mathrm{f}}$ is maximum when $\mathrm{i} \times \mathrm{m}$ is maximum.
- $\mathrm{m}_{1}=\frac{180}{60}=3, \mathrm{i}=1+\alpha$
Hence
$\Delta \mathrm{T}{\mathrm{f}}=(1+\alpha) \cdot \mathrm{k}{\mathrm{f}}=3 \times 1.86=5.58^{\circ} \mathrm{C}(\alpha<1)$
-
$\mathrm{m}{2}=\frac{180}{60}=3, \mathrm{i}=0.5, \Delta \mathrm{T}{\mathrm{f}}=\frac{3}{2} \times \mathrm{k}_{\mathrm{f}}^{\prime}=7.68^{\circ} \mathrm{C}$
-
$\mathrm{m}{3}=\frac{180}{122}=1.48, \mathrm{i}=0.5, \Delta \mathrm{T}{\mathrm{f}}=\frac{1.48}{2} \times \mathrm{k}_{\mathrm{f}}^{\prime}=3.8^{\circ} \mathrm{C}$
-
$\mathrm{m}{4}=\frac{180}{180}=1, \mathrm{i}=1, \Delta \mathrm{T}{\mathrm{f}}=1 \cdot \mathrm{k}_{\mathrm{f}}^{\prime}=1.86^{\circ} \mathrm{C}$
As per NCERT, $\mathrm{k}{\mathrm{f}}{ }^{\prime}\left(\mathrm{H}{2} \mathrm{O}\right)=1.86 \mathrm{k} \cdot \mathrm{kgmol}^{-1}$
$\mathrm{k}_{\mathrm{f}}{ }^{\prime}($ Benzene $)=5.12 \mathrm{k} \cdot \mathrm{kgmol}^{-1}$
