Question

A solution of urea (mol. mass 56 $gmo{l}^{-1}$) boils at ${100.18}^{o}C$ at the atmospheric pressure. If $K_f$ and $K_b$ for water are 1.86 and 0.52 $Kkg.mo{l}^{-1}$ respectively, the above solution will freeze at:

• A. $-{6.54}^o$
• B. ${6.54}^{o}C$
• C. ${0.654}^{o}C$
• D. $-{0.654}^{o}C$

Answer 1

The correct option is D $-{0.654}^{o}C$

$\because \Delta T_f=K_f\times$ molality of solution

and $\Delta T_b=K_b\times$ molality of solution

$\frac{\Delta T_f}{\Delta T_b}=\frac{K_f}{K_b}$

Given that

$\Delta T_b=T_2-T_1=100.18-100={0.18}^{o}C$

$K_f$ for water $=$ 1.86 K kg mol${}^{-1}$

$K_b$ for water $=$ 0.52 K log mol${}^{-1}$

$\therefore \frac{\Delta T_f}{0.18}=\frac{1.86}{0.512}$

or $\Delta T_f=\frac{1.86\times 0.18}{0.512}$

$=0.6539\approx 0.65$

$\Delta T_f=T_1-T_2$

$0.654=0^{o}C-T_2$

$\therefore T_2=-{0654}^{o}C$

($T_2⟶$ Freezing point of aqueous urea solution).