Question

Calculate the freezing point expected for $0.0711m$ aqueous solution of $N{a_2}S{O_4}$ . If this solution actually freezes at $ - {0.320^ \circ }C$ , what would be the value of the Van’t Hoff factor?
( ${K_f}$ for water is given as $1.86K\,Kg\,mo{l^ - }$ )

Answer 1

Hint:
Van’t Hoff factor is used to determine ratio in properties which are affected by the number of solute particles. Freezing point is one such property which will be affected. Since, the solvent used is an aqueous medium, the initial freezing point used will be water.

Formula used:
1. $\Delta {T_f} = m{K_f}$
Where: $\Delta T{_f}$ is the difference in freezing point.
$m$ is the molality
${K_f}$ is the cryoscopic constant
2$i=\dfrac{{observed\,value\,for\,freezing\,po\operatorname{int} }}{{theoretical\,value\,for\,freezing\,po\operatorname{int} }}$

Complete answer:
Let us first understand what Van’t Hoff factor means.
We dissolve a solute in solvent, we assume that there is no alteration to the chemical species and they exist as two independent units. However, this is not the case in reality.
Solutes undergo physical and sometimes chemical changes in the solvent , hence there is an interaction between the solute and solvent.
For example, the solute can undergo association of disassociation when it is dissolved in a solvent.
Hence, the theoretical values we had calculated for them, for example their molecular mass, will not be the same as when the molecular mass is calculated practically.
Hence, we can say Van’t Hoff factor tells us the difference between the theoretical value and observed values.
The formula we will be using today is:
$\Delta {T_f} = m{K_f}$
Where,
$\Delta T{_f}$ is the difference in freezing point, also called the depression in freezing point.
$m$is the molality
${K_f}$ is the cryoscopic constant ( it’s value has been given in the question)
Since, the solution specifies the medium is aqueous , hence the freezing point of water will be used which is ${0^ \circ }C$ .
$\Delta {T_f} = {T^ \circ } - {T_{}}$
Where, ${T^ \circ } = 273K({0^ \circ }C)$ which is the initial freezing point
$T$is the new freezing point, which needs to be calculated.
Given ,
$m = 0.071,{K_f} = 1.81$
Substituting these values, we get:
$\Delta {T_f} = 0.071 \times 1.81 \Rightarrow 0.132K$
Now, as mentioned above:
$\Delta {T_f} = {T^ \circ } - {T_{}}$
And we know: ${T^ \circ } = 273K({0^ \circ }C)$
Substituting this value we get:
$0.132 = 0 - {T_{}}$
Solving this, we get
$T = - {0.132^ \circ }C$

Now, the formula for Van’t Hoff factor is:
$i = \dfrac{{observed\,value\,for\,freezing\,po\operatorname{int} }}{{theoretical\,value\,for\,freezing\,po\operatorname{int} }}$
Where, $i$ is the Van’t Hoff factor.
Observed value $ = - 0.320$
Theoretical Value $ = - 0.132$
Substituting these values we get:
$i = \dfrac{{ - 0.320}}{{ - 0.132}} \Rightarrow 2.42 \sim 2$
Hence the value for Van’t Hoff factor is two.

Note: Depression in freezing point refers to the lowering of freezing point of the solvent when a solute is added into it. When a non-volatile solute is added into the solvent, its vapour pressure decreases, and hence, it freezes at a different temperature.