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: $\text{Number of moles (mol)}={\displaystyle \frac{\text{Mass (g)}}{\text{Molar mass (g mo}{\text{l}}^{-1})}}$
$\text{Molality (mol k}{\text{g}}^{-1}\text{)}={\displaystyle \frac{\text{Number of moles of solute (mol)}}{\text{Mass of solvent (kg)}}}$
$\mathrm{\Delta }{T}_b=K_b\times m$

Complete answer:
Calculate the number of moles of glucose using the equation as follows:
$\text{Number of moles (mol)}={\displaystyle \frac{\text{Mass (g)}}{\text{Molar mass (g mo}{\text{l}}^{-1})}}$
Substitute $\text{18 g}$ for the mass of glucose, $180\text{ g mo}{\text{l}}^{-1}$ for the molar mass of glucose. Thus,
$\text{Number of moles of glucose}={\displaystyle \frac{\text{18 g}}{180\text{ g mo}{\text{l}}^{-1}}}$
$\text{Number of moles of glucose}=0.1\text{ mol}$
Thus, the number of moles of glucose are $0.1\text{ mol}$.
Calculate the molality of the solution using the equation as follows:
$\text{Molality (mol k}{\text{g}}^{-1}\text{)}={\displaystyle \frac{\text{Number of moles of solute (mol)}}{\text{Mass of solvent (kg)}}}$
Substitute $0.1\text{ mol}$ for the number of moles of glucose, $\text{1 kg}$ for the mass of water. Thus,
$\text{Molality}={\displaystyle \frac{\text{0}\text{.1 mol}}{\text{1 kg}}}$
$\text{Molality}=\text{0}\text{.1 mol k}{\text{g}}^{-1}$
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:
$\mathrm{\Delta }{T}_b=K_b\times m$
Where, $\mathrm{\Delta }{T}_b$ 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.