Chemistry - Liquid Solution Question with Solution | TestHub

The depression in freezing point of the solution can be written as follows,

${\mathrm{\Delta T}}_{\mathrm{f}}=\mathrm{i}·{\mathrm{K}}_{\mathrm{f}}·\mathrm{m}$

molality (m) $=\frac{\mathrm{w}/\mathrm{w}\%}{\mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{solute}}\times \frac{1000}{\mathrm{mass} \mathrm{of} \mathrm{solvent} \mathrm{in} \mathrm{grams}}$

$\Rightarrow {\mathrm{\Delta T}}_{\mathrm{f}}=2.67\times 1.8\times \frac{38}{98}\times \frac{1000}{62}$

$\Rightarrow {\mathrm{\Delta T}}_{\mathrm{f}}=30.05$

$\therefore$ Freezing point of solution at which battery freeze$=243 \mathrm{K}$