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

Part (a): Calculating the freezing point of glucose

For sucrose

Mass of sucrose $\left(W\right)=10g$

Mass of water $=90g$

Molecular weight of sucrose $=342g$ $mo{l}^{-1}$

Molecular weight of water $=18g$ $mo{l}^{-1}$

$\Delta T_f=K_{f}m$

$\Delta T_f=T_f\left(solvent\right)-T_f\left(solution\right)$

$\Delta T_f=273.15-269.15=4$

$m=\frac{10}{90}\times \frac{1000}{342}=\frac{1000}{3078}$

$m=0.325$

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, $T\left(solvent\right)=273.15-7.6=265.55K$

Part (b): Defining the terms

Molality: Molality is defined as the number of moles of solute dissolved in 1 kg or 1000 g of the solvent.

Abnormal molar mass: It is the molar mass obtained when the substance undergoes association or dissociation in the solution.