The aqueous solution of urea has a freezing point of $-0.6\,^\circ C$. To prepare such a solution,how many grams of urea are needed to dissolve in $3\,kg$ of water? $(M = 60\,g\,mol^{-1}, K_{f} = 1.5\,^\circ C\,kg\,mol^{-1})$

(72G) The depression in freezing point is given by $\Delta T_{f} = T_{f}^{\circ} - T_{f} = 0\,^\circ C - (-0.6\,^\circ C) = 0.6\,^\circ C$.
Using the formula $\Delta T_{f} = K_{f} \times m$,where $m$ is the molality:
$0.6 = 1.5 \times \frac{w_{2} \times 1000}{M_{2} \times W_{1}(g)}$.
Substituting the values: $0.6 = 1.5 \times \frac{w_{2}}{60 \times 3}$.
$0.6 = 1.5 \times \frac{w_{2}}{180}$.
$w_{2} = \frac{0.6 \times 180}{1.5} = 0.4 \times 180 = 72\,g$.
Thus,$72\,g$ of urea is required.