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Question: 18 g of glucose \({{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}\) (Molar ...

18 g of glucose C6H12O6{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}} (Molar mass =180 g/mol = 180{\text{ g/mol}}) is dissolved in 1 kg{\text{1 kg}} of water in a saucepan. At what temperature will this solution boil?
[Kb{K_b} for water =0.52 K kg mol−1 = 0.52{\text{ K kg mo}}{{\text{l}}^{ - 1}}, boiling point of pure water =373.15 K = 373.15{\text{ K}}]

Explanation

Solution

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

Complete answer:
Calculate the number of moles of glucose using the equation as follows:
Number of moles (mol)=Mass (g)Molar mass (g mol−1){\text{Number of moles (mol)}} = \dfrac{{{\text{Mass (g)}}}}{{{\text{Molar mass (g mo}}{{\text{l}}^{ - 1}})}}
Substitute 18 g{\text{18 g}} for the mass of glucose, 180 g mol−1180{\text{ g mo}}{{\text{l}}^{ - 1}} for the molar mass of glucose. Thus,
Number of moles of glucose=18 g180 g mol−1{\text{Number of moles of glucose}} = \dfrac{{{\text{18 g}}}}{{180{\text{ g mo}}{{\text{l}}^{ - 1}}}}
Number of moles of glucose=0.1 mol{\text{Number of moles of glucose}} = 0.1{\text{ mol}}
Thus, the number of moles of glucose are 0.1 mol0.1{\text{ mol}}.
Calculate the molality of the solution using the equation as follows:
Molality (mol kg−1)=Number of moles of solute (mol)Mass of solvent (kg){\text{Molality (mol k}}{{\text{g}}^{ - 1}}{\text{)}} = \dfrac{{{\text{Number of moles of solute (mol)}}}}{{{\text{Mass of solvent (kg)}}}}
Substitute 0.1 mol0.1{\text{ mol}} for the number of moles of glucose, 1 kg{\text{1 kg}} for the mass of water. Thus,
Molality=0.1 mol1 kg{\text{Molality}} = \dfrac{{{\text{0}}{\text{.1 mol}}}}{{{\text{1 kg}}}}
Molality=0.1 mol kg−1{\text{Molality}} = {\text{0}}{\text{.1 mol k}}{{\text{g}}^{ - 1}}
Thus, the molality of the solution is 0.1 mol kg−1{\text{0}}{\text{.1 mol k}}{{\text{g}}^{ - 1}}.
Calculate the boiling point of the solution using the relation as follows:
ΔTb=Kb×m\Delta {T_b} = {K_b} \times m
Where, ΔTb\Delta {T_b} is the boiling point elevation (Boiling point of solution – Boiling point of pure solvent)
Kb{K_b} is the boiling point elevation constant,
m is the molality of the solution.
Thus,
(Boiling point of the solution−Boiling point of the pure solvent)=Kb×m\left( {{\text{Boiling point of the solution}} - {\text{Boiling point of the pure solvent}}} \right) = {K_b} \times m
Substitute 373.15 K373.15{\text{ K}} for the boiling point of pure solvent, 0.52 K kg mol−10.52{\text{ K kg mo}}{{\text{l}}^{ - 1}} for the boiling point elevation constant, 0.1 mol kg−1{\text{0}}{\text{.1 mol k}}{{\text{g}}^{ - 1}} for the molality of the solution. Thus,
(Boiling point of the solution−373.15 K)=0.52 K kg mol−1×0.1 mol kg−1\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}}
(Boiling point of the solution−373.15 K)=0.052 K\left( {{\text{Boiling point of the solution}} - 373.15{\text{ K}}} \right) = 0.052{\text{ K}}
Boiling point of the solution=0.052 K+373.15 K{\text{Boiling point of the solution}} = 0.052{\text{ K}} + 373.15{\text{ K}}
Boiling point of the solution=373.202 K{\text{Boiling point of the solution}} = 373.202{\text{ K}}
Thus, the boiling point of the solution is 373.202 K373.202{\text{ K}}.

Thus, the solution boils at the temperature 373.202 K373.202{\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.