Question

How many grams of urea $(M.W.60gmo{l}^{-1})$ should be dissolved in $100g$ of water in order to form a solution with the difference of $101^{\circ }C$ between its boiling point and freezing point?
$[K_{f\left(H_{2}O\right)}=1.86Kkgmo{l}^{-1}$ and $K_{b\left(H_{2}O\right)}=0.52Kkgmo{l}^{-1}]$

Answer 1

Answer: 2.52

$\Delta T_b=K_b\times m$
$B.P$. of solution, $T_b=100+\Delta T_b=100+K_{b}m$
$F.P$. of solution, $T_f=0-\Delta T_f=0-K_{f}m$
$T_b-T_f=\left(100+K_{b}m\right)-\left(-K_{f}m\right)$
$101=100+0.52\times m+1.86\times m$
$101-100=2.38m$
$m=\frac{1}{2.38}=0.42$
$m=0.42kgmo{l}^{-1}$
$\therefore$ Mass of urea to be dissolved in $100g$ of water
$=\frac{0.42\times 60}{1000}\times 100=2.52g$