Question

Boiling point of benzene is $353.23{\text{ K}}$. When $1.8{\text{ g}}$ of non-volatile solute is dissolved in $90{\text{ g}}$ of benzene. Then the boiling point raised to $354.11{\text{ K}}$. Given ${K_0}({\text{benzene}}) = 2.53{\text{ K kg mo}}{{\text{l}}^{ - 1}}$. Then molecular mass of non-volatile substance is:
A. $58{\text{ g mo}}{{\text{l}}^{ - 1}}$
B. $120{\text{ g mo}}{{\text{l}}^{ - 1}}$
C. $116{\text{ g mo}}{{\text{l}}^{ - 1}}$
D. $60{\text{ g mo}}{{\text{l}}^{ - 1}}$

Answer 1

Hint: The temperature at which the vapour pressure of any liquid becomes equal to the atmospheric pressure is known as the boiling point. The increase in the boiling point of a solvent when a solute is added is known as the elevation in boiling point. The elevation in boiling point can be calculated as the difference between the boiling points of the solution after adding the non-volatile solute and pure solvent.

Complete step by step answer:
Calculate the elevation in boiling point $\left( {\Delta {T_b}} \right)$ using the equation as follows:
$\Delta {T_b} = {\text{Boiling point of the solution after adding non - volatile solute}} - {\text{Boiling point of pure solvent}}$
Where, $\Delta {T_b}$ is the boiling point elevation.
Substitute $354.11{\text{ K}}$ for the boiling point of the solution after adding non-volatile solute, $353.23{\text{ K}}$ for the boiling point of pure solvent benzene. Thus,
$\Delta {T_b} = 354.11{\text{ K}} - 353.23{\text{ K}}$
$\Delta {T_b} = 0.88{\text{ K}}$
Thus, the elevation in boiling point is $0.88{\text{ K}}$.
Convert the units of mass of solvent benzene from ${\text{g}}$ to ${\text{kg}}$ using the relation as follows:
$1{\text{ g}} = 1 \times {10^{ - 3}}{\text{ kg}}$
Thus,
${\text{Mass of benzene}} = 90{\text{ }}{{\text{g}}} \times \dfrac{{1 \times {{10}^{ - 3}}{\text{ kg}}}}{{1{\text{ }}{{\text{g}}}}}$
${\text{Mass of benzene}} = 90 \times {10^{ - 3}}{\text{ kg}}$
Thus, the mass of the solvent benzene is $90 \times {10^{ - 3}}{\text{ kg}}$.
Calculate the molecular mass of the non-volatile solute using the equation as follows:
${\text{Molecular mass of non - volatile solute}} = {K_0} \times \dfrac{{{\text{Mass of solute}}}}{{{\text{Mass of solvent}}}} \times \dfrac{1}{{\Delta {T_b}}}$
Where, $\Delta {T_b}$ is the boiling point elevation,
                ${K_0}$ is the boiling point elevation constant
Substitute $2.53{\text{ K kg mo}}{{\text{l}}^{ - 1}}$ for the boiling point elevation constant of benzene, $1.8{\text{ g}}$ for the mass of the non-volatile solute, $90 \times {10^{ - 3}}{\text{ kg}}$ for the mass of the solvent benzene, $0.88{\text{ K}}$ for the elevation in boiling point. Thus,
${\text{Molecular mass of non - volatile solute}} = 2.53{\text{ }}{{\text{K}}}{\text{ }}\ {{{\text{kg}}}}{\text{ mo}}{{\text{l}}^{ - 1}} \times \dfrac{{1.8{\text{ g}}}}{{90 \times {{10}^{ - 3}}{\text{ }}\ {{{\text{kg}}}}}} \times \dfrac{1}{{0.88{\text{ }}{{\text{K}}}}}$
${\text{Molecular mass of non - volatile solute}} = 57.5{\text{ g mo}}{{\text{l}}^{ - 1}}$
Thus, the molecular mass of the non-volatile solute is $58{\text{ g mo}}{{\text{l}}^{ - 1}}$.

Thus, the correct option is (A) $58{\text{ g mo}}{{\text{l}}^{ - 1}}$.

Note:
The elevation in the boiling point of the solvent is inversely proportional to the molecular mass of the solute added to the solvent. The boiling point elevation decreases as the molar mass of the solute increases. Thus, an increase in molar mass of solute has a small effect on the boiling point.