Q92P - Chemistry: Molecular Nature Of Matter And Change

Question

Question: Which solution has the lower freezing point?

(a)  $\mathbf{11}\mathbf{.}\mathbf{0}{\mathbf{gofCH}}_{\mathbf{3}}\mathbf{OHin}\mathbf{100}\mathbf{.}\mathbf{gofH}\mathbf{2}\mathbf{Oor}\mathbf{22}\mathbf{.}\mathbf{0}{\mathbf{gofCH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OHin}\mathbf{200}\mathbf{.}\mathbf{gofH}\mathbf{2}\mathbf{O}$

(b)  $\mathbf{20}\mathbf{.}\mathbf{0}{\mathbf{gofH}}_{\mathbf{2}}\mathbf{Oin}\mathbf{1}\mathbf{.}\mathbf{00}{\mathbf{kgofCH}}_{\mathbf{3}}\mathbf{OHor}\mathbf{20}\mathbf{.}\mathbf{0}{\mathbf{gofCH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OHin}\mathbf{1}\mathbf{.}\mathbf{00}{\mathbf{kgofCH}}_{\mathbf{3}}\mathbf{OH}$

Step-by-Step Solution

Answer

Answer

 

1.  $\mathbf{11}{\mathbf{gofCH}}_{\mathbf{3}}\mathbf{OHin}\mathbf{100}\mathbf{gofH}\mathbf{2}\mathbf{O}$ will have lower freezing point than $\mathbf{22}{\mathbf{gofCH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OHin}\mathbf{200}{\mathbf{gofH}}_{\mathbf{2}}\mathbf{O}$ .
2.   $\mathbf{20}\mathbf{.}\mathbf{0}{\mathbf{gofH}}_{\mathbf{2}}\mathbf{Oin}\mathbf{1}\mathbf{.}\mathbf{00}{\mathbf{kgofCH}}_{\mathbf{3}}\mathbf{OH}$will have lower freezing point than $\mathbf{20}{\mathbf{gofCH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OHin}\mathbf{1}{\mathbf{kgofCH}}_{\mathbf{3}}\mathbf{OH}$ .

Step 1: Definition

 

The magnitude of freezing point depression is directly proportional to molality or molal concentration of solute as; ${\mathbf{\Delta T}}_{\mathbf{f}}{\mathbf{\alpha mor\Delta T}}_{\mathbf{f}}\mathbf{=}{\mathbf{K}}_{\mathbf{f}}\mathbf{m}$   .

Step 2: (a) Calculate molality of both the given solutions

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

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

 

 

 

Step 3: Freezing point depression in both the solutions

 

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.

 

Step 4: (b) Calculate molality of both the given solutio ns.

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

 

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

 

Step 5: Freezing point depression in both the solutions

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.