Explanation

(a)
Molality, m=$\frac{no.of\ moles\ of\ solute\ }{weight\ of\ solution\ in\ kg\ }$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1000 \times w_{1}}{m_{1}W_{1}}$
=$\frac{1000 \times 0.6}{60 \times 200}$
= 0.05
[$\therefore Molecular\ weight\ of\ NH_{2}CONH_{2} = 60$]
Given, $\mathrm{\Delta}T_{b} = 0.05\ \ \ \$
$\ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}T_{b}\ = K_{b} \times m$
or 0.05 =$K_{b} \times 0.05$
$\therefore\ \ \ \ \ \ \ \ \ \ K_{b} = 10\ K\ mol^{- 1}$