Question

18 g of glucose ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$ (Molar mass $ = 180{\text{ g/mol}}$) is dissolved in ${\text{1 kg}}$ of water in a saucepan. At what temperature will this solution boil?
[${K_b}$ for water $ = 0.52{\text{ K kg mo}}{{\text{l}}^{ - 1}}$, boiling point of pure water $ = 373.15{\text{ K}}$]

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.

Formulae Used: ${\text{Number of moles (mol)}} = \dfrac{{{\text{Mass (g)}}}}{{{\text{Molar mass (g mo}}{{\text{l}}^{ - 1}})}}$
${\text{Molality (mol k}}{{\text{g}}^{ - 1}}{\text{)}} = \dfrac{{{\text{Number of moles of solute (mol)}}}}{{{\text{Mass of solvent (kg)}}}}$
$\Delta {T_b} = {K_b} \times m$

Complete answer:
Calculate the number of moles of glucose using the equation as follows:
${\text{Number of moles (mol)}} = \dfrac{{{\text{Mass (g)}}}}{{{\text{Molar mass (g mo}}{{\text{l}}^{ - 1}})}}$
Substitute ${\text{18 g}}$ for the mass of glucose, $180{\text{ g mo}}{{\text{l}}^{ - 1}}$ for the molar mass of glucose. Thus,
${\text{Number of moles of glucose}} = \dfrac{{{\text{18 g}}}}{{180{\text{ g mo}}{{\text{l}}^{ - 1}}}}$
${\text{Number of moles of glucose}} = 0.1{\text{ mol}}$
Thus, the number of moles of glucose are $0.1{\text{ mol}}$.
Calculate the molality of the solution using the equation as follows:
${\text{Molality (mol k}}{{\text{g}}^{ - 1}}{\text{)}} = \dfrac{{{\text{Number of moles of solute (mol)}}}}{{{\text{Mass of solvent (kg)}}}}$
Substitute $0.1{\text{ mol}}$ for the number of moles of glucose, ${\text{1 kg}}$ for the mass of water. Thus,
${\text{Molality}} = \dfrac{{{\text{0}}{\text{.1 mol}}}}{{{\text{1 kg}}}}$
${\text{Molality}} = {\text{0}}{\text{.1 mol k}}{{\text{g}}^{ - 1}}$
Thus, the molality of the solution is ${\text{0}}{\text{.1 mol k}}{{\text{g}}^{ - 1}}$.
Calculate the boiling point of the solution using the relation as follows:
$\Delta {T_b} = {K_b} \times m$
Where, $\Delta {T_b}$ is the boiling point elevation (Boiling point of solution – Boiling point of pure solvent)
${K_b}$ is the boiling point elevation constant,
m is the molality of the solution.
Thus,
$\left( {{\text{Boiling point of the solution}} - {\text{Boiling point of the pure solvent}}} \right) = {K_b} \times m$
Substitute $373.15{\text{ K}}$ for the boiling point of pure solvent, $0.52{\text{ K kg mo}}{{\text{l}}^{ - 1}}$ for the boiling point elevation constant, ${\text{0}}{\text{.1 mol k}}{{\text{g}}^{ - 1}}$ for the molality of the solution. Thus,
$\left( {{\text{Boiling point of the solution}} - 373.15{\text{ K}}} \right) = 0.52{\text{ K kg mo}}{{\text{l}}^{ - 1}} \times {\text{0}}{\text{.1 mol k}}{{\text{g}}^{ - 1}}$
$\left( {{\text{Boiling point of the solution}} - 373.15{\text{ K}}} \right) = 0.052{\text{ K}}$
${\text{Boiling point of the solution}} = 0.052{\text{ K}} + 373.15{\text{ K}}$
${\text{Boiling point of the solution}} = 373.202{\text{ K}}$
Thus, the boiling point of the solution is $373.202{\text{ K}}$.

Thus, the solution boils at the temperature $373.202{\text{ K}}$.

Note: The boiling point of any solution increases when a non-volatile solute is added to it. Thus, the boiling point of the solution is always higher than the boiling point of the pure solvent.