Question

25 ml of an aqueous solution of KCl was found to require 20 ml of 1 M $AgN{O_3}$ solution when treated using a ${K_2}Cr{O_4}$ as indicator. Depression in freezing point of KCl solution with 100% ionisation will be :
[${K_F} = {2.0^0}mo{l^{ - 1}} kg$, molarity = molality]
a.) ${3.2^0}$
b.) ${1.6^0}$
c.) ${0.8^0}$
d.) ${5.0^0}$

Answer 1

Hint: Depression in freezing point is defined as the decrease in value of freezing point of a solvent when an impurity or non volatile solvent is added to it.
It can be calculated by the formula as-
$\Delta {T_f}$ = molality × ${K_f}$ × i
Where $\Delta {T_f}$ is the depression in freezing point
${K_f}$ is the molal depression constant
And I is the van't hoff factor
Formula used:

Complete step by step answer:
In the question, we have been given that 25 ml of KCl react with 20 ml of 1 M $AgN{O_3}$ solution.
Thus, volume of KCl = 25 ml
Molarity of KCl = M
Volume of $AgN{O_3}$ solution = 1 M
Molarity of $AgN{O_3}$ solution = 1 M
When the reaction occurred; we know that during neutralization
${M_1}{V_1} = {M_2}{V_2}$
Where ${M_1}$ is the molarity of KCl solution
${M_2}$ is the molarity of $AgN{O_3}$ solution
${V_1}$ is the volume of KCl solution
And ${V_2}$ is the volume of $AgN{O_3}$ solution
Thus, by putting the values, we can find molarity
${M_1}$× 25 = 1 × 20
Therefore, ${M_1} = \dfrac{{1×20}}{{25}}$
${M_1}$ = 0.8 M
Further, we are given that molarity is equal to molality.
Thus, molality = molarity = 0.8 molal
We have the formula for finding the depression in freezing point as-
$\Delta {T_f}$ = molality × ${K_f}$ × i
We will first find value if I which is Von’t hoff factor.
i = (1+α)
As the dissociation is 100%. So, the value of α will be 1.
Thus, i = (1+1)
i = 2
So, now we have all the variables and we can find out the value of depression in freezing point as-
$\Delta {T_f}$ = molality × ${K_f}$ × i
$\Delta {T_f}$ = 0.8 M × 2.0 × 2
$\Delta {T_f}$ = ${3.2^0}$

Thus, option a.) is the correct answer.

Note: The value of α depends upon the dissociation of the ions in the solution. If the dissociation is complete i.e. 100% then the value is 1. If the dissociation is less then value is also less.
The depression in freezing point is a colligative property and depends upon the number of particles present.