Question

Dry air was drawn through bulbs containing a solution of $40grams$ of urea in $300grams$ of water, then through bulbs containing pure water at the same temperature and finally through a tube in which pumice moistened with strong $H_{2}S{O}_4$ was kept. The water bulbs lost $0.0870grams$ and the sulphuric acid tube gained $2.036grams$. Calculate the molecular weight of urea.

• A. $M=26.9$
• B. $M=46.8$
• C. $M=53.8$
• D. $M=93.6$

Answer 1

Answer: (C)

$M=53.8$

Wt. of Solute (Urea) $=40g=W_2$
Wt. of Solvent (Water) $=300g=W_1$
Mol. wt. of Solvent $=18g/mol=M_1$

Now, we now that, Relative lowering of Vapour Pressure $=\frac{\text{Loss in weight of Solvent or Water bulbs}}{\text{Gain in weight of Sulphuric acid tube}}$

$\frac{\Delta p}{p^0}=\frac{0.087}{2.036}=0.0427$

Now, Using Raoult's Law:

$M_2=\frac{W_2}{W_1}\times M_1\times \frac{p^0}{\Delta p}=\frac{40}{300}\times 18\times \frac{1}{0.0427}=53.7g/mol$