Question

Given, $0.10$ M solution of a monoprotic acid is $5\% $ ionized. What is the freezing point of the solution? The mol.wt. of the acid is $300$ and ${K_f}({H_2}O) = 1.86\,C{m^{ - 1}}$:
(a).\[{\mathbf{0}}{\text{ }} - {\text{ }}{\mathbf{189}}\,{\mathbf{C}}\]
(b).\[0{\text{ }} - {\text{ }}194\,C\;\]
(c). \[ - {\text{ }}0.199C\;\]
(d). None of these

Answer 1

Hint: ${K_f}$ is the freezing point depression constant, freezing point depression is the difference in temperature between the freezing point of the pure solvent and that of the solution. The freezing point is reached when the chemical potential of the pure liquid solvent reaches that of the pure solid solvent.
Option: $(c) - 0.199C$

Complete answer:
Mass of $1L$ solution $ = 1010gm$
We are given with the molarity of $0.10$ M by which we will calculate the mass of the acid:
$Molarity = \dfrac{{Numbe{r_{}}o{f_{}}moles}}{{Volume}} = \dfrac{{mass}}{{Molecula{r_{}}mass \times Volume}}$
Substituting the values of known quantities, we get,
$ \Rightarrow 0.10 = \dfrac{{mass}}{{300 \times 1}}$
Shifting the terms of equation, we get,
$ \Rightarrow mass = 300 \times 0.10$
Now, we will calculate the mass of solvent $ = 1010 - 300 \times 0.1$
Mass of solvent $ = 980gm$.
For calculating the freezing point we need molality, $Molality = \dfrac{{moles\,of\,solute}}{{1kg\,of\,solvent}}$
\[ \Rightarrow Molality = \dfrac{{0.10}}{{\left( {\dfrac{{980}}{{1000}}} \right)}}\]
\[ \Rightarrow Molality = \dfrac{{0.1 \times 1000}}{{980}}\]
\[ \Rightarrow Molality = \dfrac{{100}}{{980}}\]
Cancelling common factors in numerator and denominator, we get,
\[ \Rightarrow Molality = \dfrac{5}{{49}} = 0.102\]
Now, we will calculate the freezing point: $\Delta {T_f} = i \times {K_f} \times m$
Where, i is van’t Hoff factor, it is the number of moles of particles obtained when one mole of solute dissolves, its formula is \[(1 + \alpha )\]and m is molality
$\Delta {T_f} = (1 + \alpha ) \times {K_f} \times m$
$\Delta {T_f} = (1 + \dfrac{5}{{100}}) \times 1.86 \times 0.102$
$\Delta {T_f} = 0.199^\circ C$
As the freezing point of water is zero, the freezing point of the solution will be:
${T_f} = 0 - 0.199 =- {199^\circ }C$

Optiion C is the correct answer.

Note:
We need to change the grams into kilograms, this is a must to do in any of the questions. While calculating the molality be careful not to use the mass of the entire solution, we only need the mass of solvent in kilograms. $\Delta {T_f}$ is the freezing point depression whereas ${T_f}$ is the freezing point of the solution or solvent.