JEE MAIN - Chemistry (2024 - 30th January Evening Shift - No. 5)

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

Explanation

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

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

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