$18\text{ g}$ of a non-volatile solute is dissolved in $200\text{ g}$ of $\text{H}_2\text{O}$ freezes at $272.07\text{ K}$. Calculate the molecular mass of solute ($K_f \text{ for water} = 1.86\text{ K kg mol}^{-1}$)

$155\text{ g/mol}$

The correct answer is Option (2) → $155\text{ g/mol}$ ##

To solve this numerical we use formula:

$\Delta T_f = K_f \times m$

Here: $\Delta T_f = T_{\text{pure}} - T_{\text{solution}}$

Therefore, $\Delta T_f = 273.15 - 272.07\text{ K} = 1.08\text{ K}$

Now substituting the value of $\Delta T_f$ in the formula

$\Delta T_f = K_f \times m$

$1.08 = 1.86\text{ K kg/mol} \times m$

$m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent}} \times 1000$

$\text{or } \Delta T_f = \frac{K_f \times w_{\text{B}} \times 1000}{M_{\text{B}} \times w_{\text{A}}}$

$1.08 = \frac{1.86 \times 18 \times 1000}{M_{\text{B}} \times 200}$

$M_{\text{B}} = \frac{1.86 \times 18 \times 1000}{1.08 \times 200}$

$M_{\text{B}} = 155.0120\text{ g/mol}$