1. Tardigrade
  2. Question
  3. Chemistry
  4. Elevation in boiling point of an aqueous solution of a non-electrolyte solute is 2.01° . What is the depression in freezing point of this solution? Kb( H 2 O )=0.52° mol -1 kg Kf( H 2 O )=1.86° mol -1 kg

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Elevation in boiling point of an aqueous solution of a non-electrolyte solute is $2.01^{\circ} .$ What is the depression in freezing point of this solution?
$K_{b}\left( H _{2} O \right)=0.52^{\circ} mol ^{-1} kg$
$K_{f}\left( H _{2} O \right)=1.86^{\circ} mol ^{-1} kg$

Solutions

Solution:

$\Delta T_{b}=$ Molality $\times K_{b}$

$ \Delta T_{b} =m K_{b} $

$\Delta T_{f} =m K_{f} $

$ \therefore \frac{\Delta T_{f}}{\Delta T_{b}} =\frac{K_{f}}{K_{b}} $

$ \Delta T_{f} =\Delta T_{b} \frac{K_{f}}{K_{b}}$

$=2.01 \times \frac{1.86}{0.52}=7.17^{\circ} $