Chemistry - Liquid Solution Question with Solution | TestHub

Given:
Mass of solute, $w_2=0.35$ g
Mass of solvent, $w_1=100$ g $=0.1$ kg
Elevation in boiling point, $\Delta T_b=0.01$ K
Molal elevation constant, $K_b=0.50$ K kg mol${}^{-1}$

The formula for elevation in boiling point is:
$\Delta T_b=K_b\times m$

Substituting the expression for molality $m$:
$\Delta T_b=\frac{K_b\times w_2}{M_2\times w_1\text{ (in kg)}}$

Rearranging to solve for the molar mass of the solute $M_2$:
$M_2=\frac{K_b\times w_2}{\Delta T_b\times w_1\text{ (in kg)}}$

Substituting the given values:
$M_2=\frac{0.50\times 0.35}{0.01\times 0.1}$

$M_2=\frac{0.175}{0.001}=175{\text{ g mol}}^{-1}$

Answer: $175\mathrm{g}{\mathrm{mol}}^{-1}$