Question

The molal elevation constant for water is $0.56{\text{K kg mol}}^{-1}$. Calculate the boiling point $\left(\text{in}{}^{\circ }\text{C}\right)$ of a solution made by dissolving $6\text{g}$ of urea $\left(N{H}_{2}CON{H}_2\right)$ in $200\text{g}$ of water.

Answer 1

$\Delta T_b=K_{b}m$
Where, $K_b=\text{Molal elevation constant}=0.56{\text{K kg mol}}^{-1}$
$m=\text{molality of the solution}$

$\therefore \Delta T_b=\frac{1000K_b\times w}{m\times W}$
Given, $\text{mass of solute}\left(w\right)=6.0\text{g},\text{Mass of solvent}\left(W\right)=200\text{g},\text{Molar mass of urea}\left(m\right)=60\text{g/mol}$
$\Delta T_b=\frac{1000\times 0.56\times 6.0}{200\times 60}=0.28{}^{\circ }\text{C}$
Thus, the boiling point of solution = boiling point of pure water $+\Delta T$
$\text{Boiling point of solution}=(100{}^{\circ }\text{C}+0.28{}^{\circ }\text{C})=100.28{}^{\circ }\text{C}$