Question

A 10% by mass of sucrose in water has a freezing point of \[269.15{\text{ K}}\]. Calculate the freezing point of \[10\% \] glucose in water. If the freezing point of pure water is \[273.15{\text{ K}}\]. Given molecular mass of sucrose is \[243{\text{ g mo}}{{\text{l}}^ -1 }\] and molecular weight of glucose is \[{\text{180 g mo}}{{\text{l}}^ -1 }\]

Answer 1

Hint:Depression in freezing point can be calculated using the formula. We will calculate two equations dividing both of them and we will get the value of freezing point. Consider molal depression constant the same for both solutions.
Formula used:
\[\Delta {{\text{T}}_{\text{f}}} = {{\text{K}}_{\text{f}}} \times {\text{m}}\]
Here \[\Delta {{\text{T}}_{\text{f}}}\] is depression in freezing point, \[{{\text{K}}_{\text{f}}}\] is molal depression coefficient and m is the molality of the solution.
\[{\text{Molality }} = \dfrac{{{\text{no}}{\text{. of moles}}}}{{{\text{mass of solvent in g}}}} \times 1000\]

Complete step by step solution:
Freezing point is that the temperature at which vapour pressure of liquid becomes equal to the vapour pressure of its solid state. Now addition of non volatile solute like sucrose and glucose increase the vapour pressure of the liquid that causes decrease in the freezing point. We have been given the depression in freezing point of \[10\% \] by mass solution of sucrose. 10 g of solute is present in 100 g of solution. The mass of solvent will be \[100 - 10 = 90{\text{ g}}\]
Molality is defined as the number of moles of solute present in mass of solution.
$\Rightarrow$\[{\text{Molality }} = \dfrac{{10}}{{342 \times 90}} \times 1000\]
The depression in freezing point is will be difference between the depression in freezing point of pure water and
$\Rightarrow$\[273.15 - 269.15{\text{ K}} = {{\text{K}}_{\text{f}}} \times \dfrac{{10}}{{342 \times 90}} \times 1000\]
Similarly the depression in freezing point for glucose will be:
$\Rightarrow$\[273.15 - {\text{T}} = {{\text{K}}_{\text{f}}} \times \dfrac{{10}}{{180 \times 90}} \times 1000\]
Dividing both the equations:
$\Rightarrow$\[\dfrac{4}{{273.15 - {\text{T}}}} = \dfrac{{180}}{{342}}\]
Solving the above equation we will get:
$\Rightarrow$\[{\text{T}} = 265.55{\text{ K}}\]

Hence the depression point for glucose is \[265.55{\text{ K}}\].

Note:
Depression in freezing point is a colligative property. Being colligative property means that the property depends upon the number of moles of solute and not on the nature of the solute. Molal elevation constant is a constant that depends on the nature of solvent that is why we considered it the same for both the equation.