Chemistry - Liquid Solution Question with Solution | TestHub

$\begin{array}{cc}\Delta {\text{T}}_{\text{f}}={\text{K}}_{\text{f}}\text{m}& \text{......(1)}\\ \Delta {\text{T}}_{\text{b}}={\text{K}}_{\text{b}}\text{m}& \text{......(2)}\end{array}\}\Rightarrow \frac{\Delta {\text{T}}_{\text{f}}}{\Delta {\text{T}}_{\text{b}}}=\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}$ ...... (3)

$\Delta {\text{T}}_{\text{f}}\to$Depression in freezing point

$\Delta {\text{T}}_{\text{b}}\to$Elevation in boiling point

Boiling point of water $={100}^{\text{o}}\text{C}$

$\text{Kf} =1.86 \text{K} {\text{kg mol}}^{-1}$

Boiling point of urea in water $={100.18}^{\text{o}}\text{C}$

$\text{Kb}=0.512 \text{K} {\text{kg mol}}^{-1}$

$\Rightarrow \Delta {\text{T}}_{\text{b}}=\text{0.18}$

Freezing point of water $=0^{\text{o}}\text{C}$

Freezing point of urea in water $=-{\text{T}}^{\text{o}}\text{C}$

$\Rightarrow \Delta {\text{T}}_{\text{f}}=\text{T}$

$\Rightarrow$ from equation (3),

$\frac{\text{T}}{\text{0.18}}=\frac{\text{1.86}}{\text{0.512}}\Rightarrow \text{T}=\text{0.6539}$

$\Rightarrow$ Freezing point urea in water $=-\text{0}{\text{.654}}^{\text{o}}\text{C}$