QJEEM24Chfd6a: JEE Main 2024 Chemistry - Class 12 JEEM24Chfd6a Solution: The solution from the following with highest depression in…

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

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