Question

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

Answer 1

The depression in the freezing point of the solution is given by
$\Delta T_f=$ freezing point of water $-$ freezing point of solution
$\Delta T_f=273.15-271=2.15$ K
Molar masses of glucose and sucrose are $180$ g/mol and $342$ g/mol respectively.
$100$ g of solution will contain $5$ g of glucose or $5$ g of sucrose.
Number of moles of glucose $=\frac{5}{180}=0.028$ moles
Number of moles of sucrose $=\frac{5}{342}=0.0146$ moles
Mass of solvent = total mass of solution - mass of solute $=100-5=95$ g or $0.095$ kg
Molality of sucrose solution $=\frac{0.0146}{0.095}=0.154$ m
$K_f=\frac{\Delta T_f}{\text{molality}}=\frac{2.15}{0.154}=13.97$
For glucose solution, $\Delta T_f=K_f\times m=13.97\times 0.29=4.08$
Freezing point of $5\%$ glucose solution in water = freezing point of water - $\Delta T_f=273.15-4.08=269.07$ K