Chemistry - Liquid Solution Question with Solution | TestHub

Using the formula for depression in freezing point:

$\Delta T_f=K_f\times m$

where $m$ is the molality of the solution.

Given $\frac{\Delta T_f}{K_f}=\frac{1}{1000}$

Therefore, $m=\frac{1}{1000}$ mol/kg.

Mass of $1$ litre of water is $1$ kg.

Molality $m=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

Moles of glucose $=\frac{1}{1000}\times 1=10^{-3}$ mol.

Molar mass of glucose $(C_6{H}_{12}{O}_6)$ is $180$ g/mol.

Weight of glucose added $=10^{-3}\times 180=0.18$ gm.

Answer: $0.18$ gm