Question

0.5 molal aqueous solution of a weak acid (HX) is 20% ionized. If ${K_f}$ for water is $1.86Kkgmo{l^{ - 1}}$, the lowering in freezing point of the solution is:
A. -0.56 K
B. -1.12 K
C. 0.56 K
D. 1.12 K

Answer 1

Hint: The Van’t Hoff factor of the chemical compound is described as the ratio of particle concentration created when the compound is dissolved to the concentration of the compound by mass. It is the extent to which the compound associates or dissociates in the solution. It is denoted by i.

Complete step by step answer:
Given,
The molality of the aqueous solution of weak acid is 0.5 molal.
${K_f}$ of water is $1.86Kkgmo{l^{ - 1}}$.
Weak acids are the one which does not completely dissociate in the solution. Thus, they are referred to as weak electrolytes.
It is given that 0.5 molal aqueous solution of weak acid is 20 % ionized.

The ionization reaction of the weak acid is shown below.
$HX \to {H^ + } + {X^ - }$
$weak\;acid\;\;\;0\;\;\;\;0$
$1 - a\;\;\;\;\;\;a\;\;\;\;\;a$

The value of a is 20 %.
The Van’t Hoff factor (i) is calculated as shown below.
$i = (1 - a) + a + a$
$\Rightarrow i = 1 + a$
Substitute the value of a in the above equation.
$i = 1 + 0.2$
$\Rightarrow i = 1.2$
Therefore, the value of Van’t Hoff factor(i) is 1.2.
The lowering in freezing point is calculated by the formula as shown below.
$\Delta {T_f} = i \times {K_f} \times m$
Where,
*$\Delta {T_f}$ is a depression in freezing point.
*${K_f}$ is freezing point depression constant.
* i is the Van’t Hoff factor
*m is the molality
To calculate, the lowering in freezing point substitutes the values in the above expression.
$\Delta {T_f} = 1.2 \times 1.86 \times 0.5$
$\Rightarrow \Delta {T_f} = 1.12K$
Thus, the value for the lowering of the freezing point is 1.12K.
Therefore, the correct option is __D__.


Note:
Molality of the solution is the measurement of the concentration which is defined as the number of moles of solute dissolved in one kilogram of solvent. The molality of the solution is independent of the temperature.