Question

For $30g$ of non-volatile solute, molecular mass $154$ is dissolved in $250g$ of benzene. The boiling point of benzene ${80.1^ \circ }C$. Calculate the boiling point of solution. (Given ${K_b} = 2.61$ for benzene)

Answer 1

Hint: The temperature at which its vapour pressure becomes equal to the atmospheric pressure, is the boiling point of a liquid.
Molal boiling point elevation or ebullioscopic constant is the elevation in boiling point for a $1$ molal solution i.e., a solution containing $1$ gram mole of solute dissolved in $1000g$ of the solvent.

Complete answer:
Elevation in boiling point, $\Delta {T_b}$ may be expressed as $\Delta {T_b} = {T_b} - {T_b}^ \circ $ ……. (I)
The elevation in the boiling point $({T_b})$ of a solution is directly proportional to the molal concentration of the solution, i.e. $\Delta {T_b}{\text{ }}\alpha {\text{ m}}$
$\Delta {T_b} = {K_b}m$, here $m$is the molality of the solution and represents the moles of solute in $1kg$ of solvent and ${K_b}$ is called molal boiling point constant.
Let ${w_B}$ gram of a non-volatile solute dissolved in ${w_A}$ grams of solvent and ${M_B}$ is the molar mass of the solute.
The molality, $m$ of the solution is : $m = \dfrac{{{\text{Moles of solute}} \times {\text{1000}}}}{{wt.{\text{ of solvent}}}}$
Substituting the value of m in $\Delta {T_b} = {K_b}m$ relation we get
$\Delta {T_b} = \dfrac{{{K_b} \times {w_B} \times 1000}}{{{M_B} \times {w_A}}}$ ……….. (II)
According to the given question
Mass of the solute, ${w_B} = 30g$
Molar mass of the solute,${M_B} = 154g/mol$
Mass of solvent, ${w_A} = 250g$
Boiling point of pure benzene, ${T_b}^ \circ = {80.1^ \circ }C$
${K_b}$ Of benzene = 2.61
Substituting these values in equation (II) to get $\Delta {T_b}$,
$\Delta {T_b} = \dfrac{{2.61 \times 30 \times 1000}}{{154 \times 250}}$
$\Delta {T_b} = 2.033$
Now substituting the value of $\Delta {T_b}$ in equation (I)
${T_b} = 80.1 + 2.033$
${T_b} = {82.133^ \circ }C$
The boiling point of the solution will be ${82.133^ \circ }C$ .

Note:
Elevation in boiling point is a colligative property, as the elevation in boiling point is directly proportional to the molal concentration of the solute (i.e., number of molecules) $\Delta {T_b}{\text{ }}\alpha {\text{ }}m$ , but boiling point is not a colligative property.
The units of molal boiling point elevation constant ${K_b}$ are degree/molality, i.e. $K{\text{ }}{{\text{m}}^{ - 1}}$ or $^ \circ C{\text{ }}{{\text{m}}^{ - 1}}$ or $K{\text{ kg mo}}{{\text{l}}^{ - 1}}$ .