Question

A solution of urea (mol. mass 56gmol−1) boils at 100.18oC at the atomospheric pressure. If Kf and Kb for water are 1.86 and 0.512Kgmol−1 respectively , the above solution will freeze at :-

Answer 1

Answer:

The 'freezing point of the solution' = -0.6539 C

Explanation:

We are given that,

Mol. Mass of urea = 56 g $\mathrm{mol}^{-1}$

Boiling Point of the solution of urea = 100.18 0C.

$\mathrm{K}_{\mathrm{t}}=1.86 \mathrm{Kg} \mathrm{mol}^{-1} \quad \mathrm{K}_{\mathrm{b}}=0.512 \mathrm{Kg} \mathrm{mol}^{-1}$

$\Delta \mathrm{T}=100.18^{\circ} \mathrm{C}-100^{\circ} \mathrm{C}=0.18^{\circ} \mathrm{C}$

We know,

$\Delta \mathrm{T}=\mathrm{K}_{\mathrm{b}} \times \mathrm{Molality}$

And $\begin{aligned} \therefore \text { Molality } &=\Delta \mathrm{T} / \mathrm{K}_{\mathrm{b}} \\ &=\frac{0.18}{0.512} \end{aligned}$

                 = 0.6539 °C

Thus, The 'freezing point of the solution' = -0.6539 C