Question

A solution of sucrose (molar mass$=342g.{mol}^{-1}$) has been prepared by dissolving $68.5g$ of sucrose in $1000g$ of water. The freezing point of solution obtained will be:
($K_f$ for water$=1.86Kkg{mol}^{-1}$)

• A. ${-0.570}^{o}C$
• B. ${-0.372}^{o}C$
• C. ${-0.520}^{o}C$
• D. ${+0.372}^{o}C$

Answer 1

The correct option is B ${-0.372}^{o}C$

A solution of sucrose (molar mass $=342g{mol}^{-1}$) has been prepared by dissolving $68.5g$ of sucrose in
$1000g$ of water. The freezing point of the solution obtained will be ${-0.372}^{o}C$

The molality of sucrose solution

$m=\frac{\left(\text{weight of sucrose}\right)}{\left(\text{molecular weight of sucrose}\right)\times \left(\text{mass of water (in g)}\right)}\times \frac{\text{1000 g}}{1kg}$

$m=\frac{\left(\text{68.5 g}\right)}{\left(\text{342 g/mol}\right)\times \left(\text{1000 g}\right)}\times \frac{\text{1000 g}}{1kg}$

$m=\text{0.200 m}$
The depression in the freezing point $\Delta T_f=K_{f}m=\text{1.86 K kg/mol}\times \text{0.200 m}=\text{0.372 K}={\text{0.372}}^{o}C$.
The freezing point of water is ${\text{0}}^{o}C$.
The freezing point of solution is $={\text{0}}^{o}C-{\text{0.372}}^{o}C=-{\text{0.372}}^{o}C$.