91Ó°ÊÓ

The magnitude of freezing point depression is directly proportional to molality or molal concentration of solute as; ${\mathbf{Δ°Õ}}_{\mathbf{f}}{\mathbf{Î\pm morΔ°Õ}}_{\mathbf{f}}\mathbf{=}{\mathbf{K}}_{\mathbf{f}}\mathbf{m}$ .

$\mathbf{molality}\mathbf{=}\frac{\mathbf{molesofsolute}}{\mathbf{kgofsolvent}\left(\mathbf{massofsolvent}\right)}$

$$\begin{gathered} {\mathbf{MolalityofCH}}_{\mathbf{3}}\mathbf{OH}\mathbf{=}\left[\frac{11{\mathrm{gCH}}_3\mathrm{OH}}{32.04{\mathrm{gCH}}_3\mathrm{OH}}\right]\left[\frac{1000g}{100{\mathrm{gH}}_{2}O}\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{433}\mathbf{m} \\ {\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OH}\mathbf{=}\left[\frac{\mathbf{22}{\mathbf{gCH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OH}}{\mathbf{46}\mathbf{.}\mathbf{07}{\mathbf{gCH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OH}}\right]\left[\frac{\mathbf{1000}\mathbf{g}}{\mathbf{200}{\mathbf{gH}}_{\mathbf{2}}\mathbf{O}}\right] \\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{387}\mathbf{m} \end{gathered}$$

Freezing point depression is directly proportional to molality of the solute. Molality of ${\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}$ as calculated is 3.433m while that of${\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OH}$ is 2.387 m.

Molality of $C{H}_{3}OH+H_{2}O$ is more so, depression in freezing point will be more and hence it will have a lower freezing point.

$\mathbf{molality}\mathbf{=}\frac{\mathbf{molesofsolute}}{\mathbf{kgofsolvent}\left(\mathbf{massofsolvent}\right)}$

$$\begin{gathered} {\mathbf{MolalityofH}}_{\mathbf{2}}\mathbf{O}\mathbf{=}\left[\frac{\mathbf{20}{\mathbf{gH}}_{\mathbf{2}}\mathbf{O}}{\mathbf{18}{\mathbf{gH}}_{\mathbf{2}}\mathbf{O}}\right]\left[\frac{\mathbf{1000}\mathbf{g}}{\mathbf{1000}{\mathbf{gCH}}_{\mathbf{3}}\mathbf{OH}}\right] \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{111}\mathbf{m} \\ {\mathbf{MolalityofCH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OH}\mathbf{=}\left[\frac{\mathbf{20}{\mathbf{gCH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OH}}{\mathbf{46}\mathbf{.}\mathbf{07}{\mathbf{gCH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OH}}\right]\left[\frac{\mathbf{1000}\mathbf{g}}{\mathbf{1000}{\mathbf{gCH}}_{\mathbf{3}}\mathbf{OH}}\right] \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{434}\mathbf{m} \end{gathered}$$

Freezing point depression is directly proportional to molality of the solute. Molality of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ as calculated is 1.111 m while that of ${\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OH}$ is 0.434 m.

Molality of $H_{2}O+C{H}_{3}OH$ is more so, depression in freezing point will be more and hence it will have a lower freezing point.