Question

What will be the molar mass of ${\text{NaCl}}$ determined experimentally following elevation in the boiling point or depression in freezing point method?
A. $ < 58.5{\text{ g mo}}{{\text{l}}^{ - 1}}$
B. $ > 58.5{\text{ g mo}}{{\text{l}}^{ - 1}}$
C. $ = 58.5{\text{ g mo}}{{\text{l}}^{ - 1}}$
D. None of these

Answer 1

Hint: The boiling point of a pure solvent increases when a non-volatile solute is added to it. This increase in the boiling point is known as the elevation in boiling point. The freezing point of a pure solvent decreases when a non-volatile solute is added to it. This decrease in the freezing point is known as the depression in freezing point.

Formulae Used:
$\Delta {T_b} = {K_b} \times m \times i$
$\Delta {T_f} = {K_f} \times m \times i$
$i = \dfrac{{{\text{Experimental molecular weight}}}}{{{\text{Calculated molecular weight}}}}$

Complete step by step solution:
The equation for the elevation in boiling point of a solution is,
$\Delta {T_b} = {K_b} \times m \times i$
Where,
$\Delta {T_b}$ is the elevation in boiling point,
${K_b}$ is the constant of the elevation in boiling point,
$m$ is the molality of the solution,
$i$ is the van’t Hoff factor.
The equation for the depression in freezing point of a solution is,
$\Delta {T_f} = {K_f} \times m \times i$
Where, $\Delta {T_f}$ is the depression in freezing point,
${K_f}$ is the constant of the depression in freezing point,
$m$ is the molality of the solution,
$i$ is the van’t Hoff factor.
From the two equations, we can say that the factor that affects the molecular weight is the van’t Hoff factor. All the other quantities are constant for a solution.
${\text{NaCl}}$ dissociates into one ${\text{N}}{{\text{a}}^ + }$ ion and one ${\text{C}}{{\text{l}}^ - }$ ion i.e. total two ions. Thus, the van’t Hoff factor for ${\text{NaCl}}$ is 2.
The calculated molecular mass of ${\text{NaCl}}$ is,
${\text{Molar mass of NaCl}} = {\text{Atomic mass of Na}} + {\text{Atomic mass of Cl}}$
${\text{Molar mass of NaCl}} = {\text{23}} + 35.5$
${\text{Molar mass of NaCl}} = {\text{58}}{\text{.5}}$
Thus, the calculated molar mass of ${\text{NaCl}}$ is ${\text{58}}{\text{.5}}$.
The experimental molar mass of ${\text{NaCl}}$ is calculated using the base equation of van’t Hoff factor,
$i = \dfrac{{{\text{Experimental molar mass}}}}{{{\text{Calculated molar mass}}}}$
Thus,
${\text{Experimental molar mass}} = i \times {\text{Calculated molar mass}}$
${\text{Experimental molar mass}} = 2 \times {\text{58}}{\text{.5}}$
${\text{Experimental molar mass}} = 117$
Thus, the experimental molar mass of ${\text{NaCl}}$ is ${\text{117}}$.
Thus, the experimental molar mass is greater than the calculated molar mass.
Thus, the molar mass of ${\text{NaCl}}$ determined experimentally following elevation in the boiling point or depression in freezing point method is $ > 58.5{\text{ g mo}}{{\text{l}}^{ - 1}}$.

Thus, the correct option is (B) $ > 58.5{\text{ g mo}}{{\text{l}}^{ - 1}}$.

Note:
The van’t Hoff factor is the ratio of the experimental molar mass to the calculated molar mass. For ${\text{NaCl}}$ the van’t Hoff factor is 2.