Question

A solution is prepared by dissolving 27.0 g of urea $\mathbf{\left[}\mathbf{\right(}\mathbf{N}{\mathbf{H}}_2{\mathbf{)}}_2\mathbf{CO}\mathbf{]}$ , in 150.0 g of water. Calculate the boiling point of the solution. Urea is a nonelectrolyte.

Short answer

When 27.0 g of urea is dissolved in 150.0 g of water, the boiling point of the solution will become $\text{105.6}°\text{C}$

Step-by-step solution

Molality of solution

According to the question, it is given that

Mass of solute (urea) = 27g

Mass of solvent = 150.0 g

The molar mass of solvent = 60.0 g

${\text{K}}_{\text{b}}\text{=0.52}°\text{C/mg}$

When a solute is added to a pure solvent, there is an increase in the boiling point of the solution which is given by the relation $\Delta {\text{T}}_{\text{b}}{\text{=k}}_{\text{b}}\times \text{m}\times \text{i}$.

Here,$\Delta {\text{T}}_{\text{b}}$is the difference between the boiling point of the solution and the boiling point of the pure solution.

The molality of the solution will be

$\begin{array}{l}\text{Molality=}\frac{\text{Massofsolute}\times \text{1000}}{\text{Molarmassofsolute}\times \text{Massofsolvent}}\\ \text{Molality=}\frac{\text{27}\times \text{1000}}{\text{60}\times \text{150}}\text{=3m}\end{array}$

Boiling point of the solution

Thus,

$\begin{array}{l}\text{Boilingpointofsolution-100=0.52}\times \text{1}\times \text{3}\\ \text{Boilingpointofsolution=}1.56\text{+100=105.6}°\text{C}\end{array}$