Question

A 5 % solution (by mass) of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of 5% glucose in water if the freezing point of pure water is 273.15 K.

Answer 1

Hint: The depression in the freezing point is given by the following expression:
\[\Delta {T_f} = {K_f} \times \left( {\dfrac{w}{m} \times \dfrac{{1000}}{W}} \right)\]
Here, \[\Delta {T_f}\] is the depression in the freezing point, \[{K_f}\] is the molal depression in the freezing point constant, w is the mass of the solute, m is the molar mass of the solute and W is the mass of solvent.

Complete step by step answer:
The molar masses of glucose and sucrose are \[180{\text{ }}g/mol\] and \[342{\text{ }}g/mol\] respectively.
\[5\% \] solution means that \[5{\text{ }}grams\] of solute are dissolved in \[95{\text{ }}grams\] of solvent to prepare \[100{\text{ }}grams\] of solution. The depression in the freezing point is the difference between the freezing point of pure solvent (water) and the freezing point of the solution.
The depression in the freezing point is given by the following expression:
\[\Delta {T_f} = {K_f} \times \left( {\dfrac{w}{m} \times \dfrac{{1000}}{W}} \right)\]
Substitute values for cane sugar:
\[\left( {273.15{\text{ K }} - {\text{ 271 K}}} \right) = {K_f} \times \left( {\dfrac{5}{{342}} \times \dfrac{{1000}}{{95}}} \right)\]… …(1)
Substitute values for glucose:
\[\Delta {T_f} = {K_f} \times \left( {\dfrac{5}{{180}} \times \dfrac{{1000}}{{95}}} \right)\]… …(2)
Divide equation (2 with equation (1)
\[\dfrac{{\Delta {T_f}}}{{2.15}} = \dfrac{{342}}{{180}}\]
Rearrange above equation:
\[
\Delta {T_f} = \dfrac{{342}}{{180}} \times 2.15 \\
= 4.09{\text{ K}} \\
\]
Subtract the depression in the freezing point of 5 % glucose solution from the freezing point of water to obtain the freezing point of 5 % glucose solution:
\[{\text{273}}{\text{.15 K}} - {\text{4}}{\text{.09 K = 269}}{\text{.06 K}}\]
Hence, the freezing point of 5% glucose solution is \[269.06{\text{ }}K\] .

Note: The depression in the freezing point is the colligative property. It depends on the number of solute particles and is independent of the nature of the solute. The number of solute particles present in 5 % solution of sucrose is different from the number of solute particles present in 5 % solutions of glucose because the molar masses of sucrose and glucose are different. Equal masses of glucose and sucrose samples will have different numbers of moles as they have different molar masses.