Question

A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.
(Given: Molar mass of sucrose = 342$g\ \text{mo}{{\text{l}}^{\text{-1}}}$, molar mass of glucose = 180 $g\ \text{mo}{{\text{l}}^{\text{-1}}}$)
A. 275.53 K
B. 265.53 K
C. 271.32 K
D. 282.43 K

Answer 1

Hint: Freezing defines with the phase change of a substance from liquid state to solid state i.e. transformation of one state to other as we know there are three states of matter solid, liquid and gas so freezing defines change from liquid to solid.

Complete step-by-step answer:Freezing point is defined as the temperature of a liquid which changes its state from liquid to solid at atmospheric pressure and at this point the two phases i.e. solid and liquid are in equilibrium with each other.
Freezing point can be calculated by the formula:
$\Delta {{T}_{f}}={{K}_{f}}m$where ${{T}_{f}}=265.55K$ $\Delta {{T}_{f}}$= change in freezing point, m is molality and ${{K}_{f}}$is molal freezing point depression point.
For sucrose, $\Delta {{T}_{f}}={{K}_{f}}\times \dfrac{10}{342}\times \dfrac{1}{Mass\ \text{of}\ \text{solvent}}$
For glucose, $\Delta {{T}_{f}}={{K}_{f}}\times \dfrac{10}{180}\times \dfrac{1}{Mass\ \text{of}\ \text{solvent}}$
Here both have same mass of solvent so we can calculate the value by dividing the both equations:
$\dfrac{\Delta {{T}_{f}}(sucrose)}{\Delta {{T}_{f}}(glucose)}=\dfrac{180}{342}$
Now change in freezing point in this can be calculated as follows:
$\dfrac{{{T}_{f}}^{{}^\circ }-{{T}_{f}}(sucrose)}{{{T}_{f}}^{{}^\circ }-{{T}_{f}}(glucose)}=\dfrac{180}{342}$
$\Rightarrow \dfrac{273.15-269.15}{273.15-{{T}_{f}}(glucose)}=\dfrac{180}{342}$
$\Rightarrow \dfrac{4}{273.15-{{T}_{f}}(glucose)}=0.526$
$\Rightarrow {{T}_{f}}=273.15-\dfrac{4}{0.526}$
$\therefore {{T}_{f}}=265.55K$

Hence the value of freezing point of glucose in water is 265.55 K, option B is the correct answer.

Note: Freezing point is affected by temperature as at high temperature freezing point is higher as at this temperature molecules are closely packed with strong intermolecular forces but as the temperature decreases forces also get weakened and molecules will turn into liquid.