Question

A 0.1 molal aqueous solution of a weak acid is 30 % ionized. If ${{K}_{f}}$ for water is ${{1.86}^{\circ }}C/m$, the freezing point of the solution will be:
(A) $-{{0.18}^{\circ }}C$
(B) $-{{0.54}^{\circ }}C$
(C) $-{{0.36}^{\circ }}C$
(D) $-{{0.24}^{\circ }}C$

Answer 1

As we know that freezing is the phase change of a substance from a liquid state to a solid-state. We have to find the freezing point ($\Delta {{T}_{f}}$) of solution which is the temperature of a liquid at which it changes its state from liquid to solid at atmospheric pressure. To find the freezing point, formula used is-$\Delta {{T}_{f}}=i.{{K}_{f}}.m$

Complete step by step answer:
- We are provided with a molality of aqueous solution (m) = 0.1 molal
And it is said that it is 30 % ionized, therefore the degree of dissociation will be 0.3
- As it is a weak acid, therefore, it will break in such a way-
$HA\to {{H}^{+}}+{{A}^{-}}$

Initial	1	0	0
Final	1-$\alpha$	$\alpha$	$\alpha$.

From the above reaction, we can see that if initially, HA is 1mole so ${{H}^{+}}$ and ${{A}^{-}}$ both are 0 moles. then after sometime HA will become $1-\alpha$ and ${{H}^{+}}$, as well as ${{A}^{-}}$, will become $\alpha$.
Now if we have to write Vant Hoff factor(i) , it is = $\dfrac{total\text{ }moles}{initial\text{ }moles}$
We have total moles = $1-\alpha +\alpha +\alpha$
Initial moles = 1
So vant Hoff factor(i) =

& \dfrac{1-\alpha +\alpha +\alpha }{1} \\\ & =1+\alpha \\\ & =1 + 0.3 \\\ & =1.3 \\\ \end{aligned}$$ \- Therefore i=1.3 Here , ${{K}_{f}}$ is the freezing point constant, Now, to find the freezing point -$\Delta {{T}_{f}}=i.{{K}_{f}}.m$ $\begin{aligned} & \Delta {{T}_{f}}=1.3\times 1.86\times 0.1 \\\ & =0.241 \\\ \end{aligned}$ -We have the formula of freezing point of solution, that is- ${{T}_{F}}S={{T}_{F}}^{\circ }-\Delta {{T}_{F}}$ Where, ${{T}_{F}}^{\circ }$ is the freezing point of water, which is equals to ${{0}^{\circ }}C$ Now, $\begin{aligned} & {{T}_{F}}S={{0}^{\circ }}C-{{0.241}^{\circ }}C \\\ & =-{{0.241}^{\circ }}C \\\ \end{aligned}$ Hence, we can conclude that correct option is (C) ,that is the freezing point of solution will be- $-{{0.241}^{\circ }}C$. **So, the correct answer is “Option C”.** **Additional Information”.** As we know that there are the main factors that affect the freezing point is the type of molecules the liquid is made up of. Let’s see some of them- \- If intermolecular forces between its molecules are strong then there will be a high freezing point. if the forces are weak the freezing point is relatively low. \- as we know that theoretically, the freezing point and the melting point of any substance should be the same, and for many substances, slight differences have been observed in these two temperatures. It is because substances can be cooled below their freezing point. **Note:** One must not confuse between the terms molal and molar. Herein this particular question, 0.1molal aqueous solution is given which is the unit of molality. And molar is the unit of molarity. \- Molarity is the ratio of the moles of solute to the total liters of solution. \- Molality is the ratio of moles of solute to the kilograms of a solvent.