Question

On dissolving 0.5 g of non-volatile, non-ionic solute to 39 g of benzene, its vapour pressure decreases from 650 mm of Hg to 640 mm of Hg. The depression of freezing point of benzene (in K) upon addition of the solute is __________________. [Given data: Molar mass & molar freezing point depression of benzene is $78\mathrm{g}{\mathrm{mol}}^{-1}$ & $5.12\mathrm{K}\mathrm{Kg}{\mathrm{mol}}^{-1}$]

Answer 1

Step 1: Given data

Molar mass of benzene is = $78\mathrm{g}{\mathrm{mol}}^{-1}$

Molar freezing point depression of benzene is = $5.12\mathrm{K}\mathrm{Kg}{\mathrm{mol}}^{-1}$

Mass of Benzene given is $39\mathrm{g}$

Mass of Solute given is $0.5\mathrm{g}$

Initial vapour pressure is $650\mathrm{nm}$ of Hg

Final vapour pressure is $640\mathrm{nm}$ of Hg

Step 2: Calculating the moles of solute

Let the mass of salute be ${\mathrm{M}}_{\mathrm{solute}}$

We have,

$$\begin{gathered} \frac{{\mathrm{P}}^0-{\mathrm{P}}_{\mathrm{S}}}{{\mathrm{P}}_{\mathrm{S}}}=\mathrm{i}\times \left[\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\mathrm{n}}_{\mathrm{solvent}}}\right] \\ \Rightarrow \frac{650-640}{650}=\mathrm{i}\times \left[\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\displaystyle \frac{\mathrm{Given}\mathrm{weight}\mathrm{of}\mathrm{benzene}}{\mathrm{molecular}\mathrm{weight}\mathrm{of}\mathrm{benzene}}}}\right] \\ \Rightarrow \frac{650-640}{650}=1\times \left[\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\displaystyle \frac{39}{78}}}\right] \\ \Rightarrow \frac{10}{650}\times \frac{39}{78}={\mathrm{n}}_{\mathrm{solute}} \\ \Rightarrow {\mathrm{n}}_{\mathrm{solute}}=\frac{0.5}{64} \end{gathered}$$

Step 3: Calculating the depression of freezing point of benzene

Molality is calculated by,

$$\begin{gathered} \mathrm{molality}=\frac{\mathrm{no}.\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{the}\mathrm{solute}}{\mathrm{weight}\mathrm{of}\mathrm{solvent}\mathrm{in}kgs} \\ \Rightarrow \mathrm{molality}=\frac{{\mathrm{n}}_{\mathrm{solute}}}{\mathrm{weight}\mathrm{of}\mathrm{solvent}\mathrm{in}kgs} \end{gathered}$$

We know that,

$$\begin{gathered} \mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m} \\ \mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \frac{{\mathrm{n}}_{\mathrm{solute}}}{\mathrm{weight}\mathrm{of}\mathrm{solvent}}\begin{array}{}\end{array} \end{gathered}$$

Here $\mathrm{\Delta }{T}_f$ is the depression in freezing point, $i$ is the Van’t Hoff factor, $K_f$ is the cryoscopic constant, and $m$ is the molality

$\therefore$By putting the values we get,

$$\begin{gathered} \mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{m} \\ \mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \frac{{\mathrm{n}}_{\mathrm{solute}}}{\mathrm{weight}\mathrm{of}\mathrm{solvent}} \\ \Rightarrow \mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}=1\times 5.12\times \frac{{\displaystyle \frac{0.5}{64}}}{39\times {10}^{-3}} \\ \Rightarrow \mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}=5.12\times \frac{0.0078}{39}\times {10}^3 \\ \Rightarrow \mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}=5.12\times 0.20032 \\ \Rightarrow \mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}=1.025\mathrm{K} \\ \begin{array}{}\end{array} \end{gathered}$$

Therefore, The depression of freezing point of benzene (in K) upon addition of the solute is $∆{\mathrm{T}}_{\mathrm{f}}=1.025\mathrm{K}$.