Question

Which of the following aqueous solutions has the highest freezing point?
A) $0.1\,m$ Sucrose
B) $0.01\,m\,NaCl$
C) $0.1\,m\,NaCl$
D) $0.01\,m\,N{{a}_{2}}S{{O}_{4}}$

Answer 1

Hint: Van't Hoff factor is directly proportional to change in depression in freezing point but inversely proportional to the freezing point. For ionic compounds when dissolved in aqueous solution, the Van't Hoff factor is equal to the number of dissociated ions in a reaction of a substance.

Formula used: $\Delta {{T}_{f}}=i{{K}_{f}}m$
where, $\Delta {{T}_{f}}$ is the change in depression in freezing point, $i$ is the Vant Hoff factor, ${{K}_{f}}$ is the freezing point constant and $m$ is the molality.

Complete step-by-step answer:
Let us write the formula for freezing point
$\Delta {{T}_{f}}=i{{K}_{f}}m$
where, $\Delta {{T}_{f}}$ is the change in depression in freezing point, $i$ is the Van't Hoff factor, ${{K}_{f}}$ is the freezing point constant and $m$ is the molality.
Change in depression freezing point is directly proportional to Van't Hoff factor and molality.

Now, let us discuss the given options here, one by one.
A) $0.1\,m$ Sucrose
Here, the molality of Sucrose is $0.1\,m$
Sucrose is a non – electrolyte which does not dissociate or associate in aqueous solution.
$i=1+(n-1)\alpha $
where, $\alpha $ is the degree of dissociation.
As we have discussed that nonelectrolyte shows no dissociation or association, therefore the degree of dissociation is zero. $(\alpha =0)$
Therefore, $i=1$
$\Delta {{T}_{f}}=i{{K}_{f}}m$
 Now, substituting the values of Van't Hoff factor and molality in the above formula, we get
$\Rightarrow \Delta {{T}_{f}}=1\times {{K}_{f}}\times 0.1$
$\Rightarrow \Delta {{T}_{f}}=0.1{{K}_{f}}$

B) $0.01\,m\,NaCl$
Here, the molality of $NaCl$ is $0.01\,m$
$NaCl\to N{{a}^{+}}+C{{l}^{-}}$
Here, Van't Hoff factor, $i=2$
$\Delta {{T}_{f}}=i{{K}_{f}}m$
Now, substituting the values of Van't Hoff factor and molality in the above formula, we get
$\Rightarrow \Delta {{T}_{f}}=2\times {{K}_{f}}\times 0.01$
$\Rightarrow \Delta {{T}_{f}}=0.02{{K}_{f}}$

C) $0.1\,m\,NaCl$
Here, the molality of $NaCl$ is $0.1\,m$
$NaCl\to N{{a}^{+}}+C{{l}^{-}}$
Here, Van't Hoff factor, $i=2$
$\Delta {{T}_{f}}=i{{K}_{f}}m$
 Now, substituting the values of Van't Hoff factor and molality in the above formula, we get
$\Rightarrow \Delta {{T}_{f}}=2\times {{K}_{f}}\times 0.1$
$\Rightarrow \Delta {{T}_{f}}=0.2{{K}_{f}}$

D) $0.01\,m\,N{{a}_{2}}S{{O}_{4}}$
Here, the molality of $N{{a}_{2}}S{{O}_{4}}$ is $0.01\,m$
\[N{{a}_{2}}S{{O}_{4}}\to 2N{{a}^{+}}+S{{O}_{4}}^{2-}\]
Here, Van't Hoff factor, $i=3$
$\Delta {{T}_{f}}=i{{K}_{f}}m$
Now, substituting the values of Van't Hoff factor and molality in the above formula, we get
$\Rightarrow \Delta {{T}_{f}}=3\times {{K}_{f}}\times 0.01$
$\Rightarrow \Delta {{T}_{f}}=0.03{{K}_{f}}$
Van't Hoff factor is directly proportional to change in depression in freezing point but inversely proportional to the freezing point.
In this question, we can see that the least value of the Van't Hoff factor is $0.01\,m\,NaCl$ , therefore it shows the highest freezing point.

Hence, the correct option is (B) $0.01\,m\,NaCl$

Note: Non – electrolytes are defined as the compounds that do not ionize in aqueous solution. They do not conduct electricity. They are held together by a covalent bond. For example, sucrose, glucose. When a certain solute is added to a solvent, it decreases the freezing point of the solution, and hence it is known as freezing point depression.