Question

\[0 \cdot 7{\text{ g}}\] of an organic compound when dissolved in $32{\text{ g}}$ of acetone produces an elevation of \[0 \cdot {25^ \circ }{\text{C}}\] in the boiling point. Calculate the molecular mass of organic compound (${K_b}$ for acetone $ = 1 \cdot 72{\text{ K kg mol}}{{\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. Thus, the formula for the elevation in boiling point is,
$\Delta {T_b} = {K_b} \times m$
Where, $\Delta {T_b}$ is the boiling point elevation,
${K_b}$ is the boiling point elevation constant,
$m$ is the molality of the solution
Thus,
${\text{Molar mass of solute}} = {K_b} \times \dfrac{{{\text{Mass of solute}}}}{{{\text{Mass of solvent}}}} \times \dfrac{1}{{\Delta {T_b}}}$
Where, $\Delta {T_b}$ is the boiling point elevation,
 ${K_b}$ is the boiling point elevation constant

Complete step by step answer:
Step 1: Convert the units of the boiling point elevation from $^ \circ {\text{C}}$ to ${\text{K}}$ as follows:
The boiling point elevation is $0 \cdot {25^ \circ }{\text{C}}$. The boiling point elevation in the unit of ${\text{K}}$ has the same value.
Thus, the boiling point elevation is $0 \cdot 25{\text{ K}}$.
Step 2: Convert the units of mass of solvent acetone from ${\text{g}}$ to ${\text{kg}}$ using the relation as follows:
$1{\text{ g}} = 1 \times {10^{ - 3}}{\text{ kg}}$
Thus,
${\text{Mass of solvent}} = 32\not{{\text{g}}} \times \dfrac{{1 \times {{10}^{ - 3}}{\text{ kg}}}}{{1\not{{\text{g}}}}}$
${\text{Mass of solvent}} = 0 \cdot 032{\text{ kg}}$
Thus, the mass of the solvent acetone is $0 \cdot 032{\text{ kg}}$
Step 3: Calculate the molecular mass of the organic compound using the equation as follows:
${\text{Molar mass of solute}} = {K_b} \times \dfrac{{{\text{Mass of solute}}}}{{{\text{Mass of solvent}}}} \times \dfrac{1}{{\Delta {T_b}}}$
Substitute $1 \cdot 72{\text{ K kg mo}}{{\text{l}}^{ - 1}}$ for the boiling point elevation constant, $0 \cdot 7{\text{ g}}$ for the mass of the organic compound, $0 \cdot 032{\text{ kg}}$ for the mass of the solvent acetone, $0 \cdot 25{\text{ K}}$ for the boiling point elevation. Thus,
${\text{Molar mass of solute}} = 1 \cdot 72\not{{\text{K}}}\not{{{\text{kg}}}}{\text{ mo}}{{\text{l}}^{ - 1}} \times \dfrac{{0 \cdot 7{\text{ g}}}}{{0 \cdot 032\not{{{\text{kg}}}}}} \times \dfrac{1}{{0 \cdot 25\not{{\text{K}}}}}$
${\text{Molar mass of solute}} = 150 \cdot 5{\text{ g mo}}{{\text{l}}^{ - 1}}$
Thus, the molecular mass of the organic compound is $150 \cdot 5{\text{ g mo}}{{\text{l}}^{ - 1}}$.

Note:
The elevation in boiling point of 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.