Question

When $0.2$ mole of urea is dissolved in $100\, g$ of a solvent, the depression of freezing point is $0.50^{\circ} C$. When $0.3$ mole of non-volatile solute is dissolved in $100\, g$ of the same solvent, the depression of freezing point will be_____ ${}^{\circ}C$

Answer 1

For the same solvent, the value of $K_{f}$ remains the same.
$\left(\Delta T _{ f }\right)_{1}= K _{ f } m _{1}$
$\left(\Delta T _{ f }\right)_{2}= K _{ f } m _{2}$
$K _{ f }=\frac{\left(\Delta T _{ f }\right)_{1}}{ m _{1}}$
and $K _{ f }=\frac{\left(\Delta T _{ f }\right)_{2}}{ m _{2}}$
$\therefore \frac{\left(\Delta T _{ f }\right)_{2}}{ m _{2}}=\frac{\left(\Delta T _{ f }\right)_{1}}{ m _{1}}$
$\therefore \left(\Delta T _{ f }\right)_{2}=\frac{\left(\Delta T _{ f }\right)_{1}}{ m _{1}} \times m _{2}$
$\therefore \left(\Delta T _{ f }\right)_{2}=\frac{0.50}{\left(\frac{0.2}{0.100}\right)} \times\left(\frac{0.3}{0.100}\right)=\frac{0.50 \times 3}{2}$
$=0.75^{\circ} C$