Question

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water $\left( {\Delta {T_f}} \right)$ when $0.01$ mol of sodium sulphate is dissolved in $1$ kg of water is:
$\left[ {{K_f} = 186K - Kg{{\left( {mol} \right)}^{ - 1}}} \right]$
A.$3.72K$
B.$5.58K$
C.$7.44K$
D.$1.86K$

Answer 1

The change in freezing point is one of the colligative properties and can be determined from the Van’t Hoff factor, molality and molal freezing point constant. Molal freezing point is constant and given, Van’t Hoff factor of sodium sulphate is $3$and the molality can be determined from the number of moles and volume of solution in kilograms.
Formula used:
$\Delta {T_f} = {K_f} \times i \times m$
$\Delta {T_f}$ is change in freezing point
${K_f}$ is molal elevation constant
$i$ is Van’t Hoff factor
$m$ is molality

Complete answer:
Colligative properties are one of the properties and change in depression freezing point is one of the colligative properties.
The number of moles of solute is $0.01$ and volume of solution is $1kg$.
Thus, the molality will be $\dfrac{{0.01}}{1} = 0.01$
The sodium sulphate is an inorganic salt and can be dissociated into three ions. Thus, Van’t Hoff factor is $3$.
The molal freezing point is already given as $186K - Kg{\left( {mol} \right)^{ - 1}}$
-Now we know, $\Delta {T_f} = {K_f} \times i \times m$
Substitute all these values in the above equation,
$\Delta {T_f} = 186 \times 3 \times 0.01 = 5.58K$
The change in freezing point of water $\left( {\Delta {T_f}} \right)$ is $5.58K$
So, option (B) is the correct answer.

Note:
The inorganic salts are the3 compounds other than carbon and hydrogen containing atoms. Sodium sulphate on complete dissociation gives two sodium ions and one sulphate ion. Thus, the van’t Hoff factor will be three but not two. The volume of solution while calculating molality should be in kilograms.