Question

The freezing point of a solution containing $4.8g$ of a compound in $60g$ of benzene is ${4.48}^{\circ }C$. What is the molar mass of the compound?
Given
$K_f\text{for benzene}=5.1K{m}^{-1}$
and freezing point of pure benzene is ${5.5}^{\circ }C$

• A. $100gmo{l}^{-1}$
• B. $200gmo{l}^{-1}$
• C. $300gmo{l}^{-1}$
• D. $400gmo{l}^{-1}$

Answer 1

The correct option is D $400gmo{l}^{-1}$

$\text{weight of solute}=4.8g$
$\text{weight of solvent}=60g$
$\text{Freezing point of solution}={4.48}^{\circ }C$
$\text{Freezing point of solvent =}{5.5}^{\circ }C$
$K_f=5.1K{m}^{-1}$
Change in Temp = Freezing point of solvent - Freezing pt.of $so{l}^n$
$\left(\Delta T_f\right)=\left(T_f^{\circ }\right)-\left(T_f\right)=({5.5}^{\circ }C-{4.48}^{\circ }C)={1.02}^{\circ }C$
$\Delta T_f=K_f\times m$
$m=\frac{1.02}{5.1}=0.2$
$m=\frac{\text{No. of moles of non- volatile solute}}{\text{Mass of solvent (in kg)}}$
$0.2=\frac{4.8\times 1000}{60\times M}\Rightarrow M=\frac{4.8\times 1000}{0.2\times 60}=400gmo{l}^{-1}$