Question

(a) 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 $mo{l}^{-1}$)
(Molar mass of glucose = 180 g $mo{l}^{-1}$)
(b) Define the following terms:
(i) Molality (m)
(ii) Abnormal molar mass.

Answer 1

(a) Mass of sucrose (W) = 10 g
Mass of water = 90 g
Molecular mass of sucrose = 342 g $mo{l}^{-1}$
Molecular weight of water = 18 g $mo{l}^{-1}$
$\Delta T_f=K_{f}m\Delta T_f=T_f$ (solvent) - $T_f$ (solution)
$\Delta T_f=273.15-269.15=4$
$m=\frac{10}{90}\times \frac{1000}{342}=\frac{1000}{3078}$
m = 0.325
So $\Delta T_f=K_f\times m$
$=\frac{4}{0.325}=12.31$
For Glucose:
$\Delta T_f=K_f\times m$
$=12.3\times \frac{10\times 1000}{180\times 90}=7.6$
$\Delta T_f=T$ (solvent) - T (Solution)
So
T (solvent) = 273.15 - 7.6
= 265.55 K
(b) (i) Molality : It is defined as the number of moles of solute dissolved in 1 kg or 1000 g of the solvent.
(ii) Abnormal molar mass : When the substance undergoes association or dissociation in the solution.