Question

To a one litre solution of 0.1 N HCl, 0.025 mol of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$ is added. Assuming 80% dissociation of the solutes, the freezing point of the solution is: (${{\text{K}}_{\text{f}}}$ = 1.85 deg/molal)
A. $- {0.33^ \circ }{\text{C}}$
B. $- {0.85^ \circ }{\text{C}}$
C. $- {0.23^ \circ }{\text{C}}$
D. $- {0.416^ \circ }{\text{C}}$

Answer 1

The freezing point of a solution is the temperature at which there is an equilibrium between the liquid and solid phase, we usually observe depression in the freezing point due to the addition of solute particles, ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$ is the solute for this question.

Complete Step by step answer:
Let’s calculate the vent off factor for the 80% dissociation of solutes,
We know, i = 1 + (n – 1)${{\alpha }}$
Here, ${{\alpha = }}\dfrac{{80}}{{100}}$
$\Rightarrow$ ${{\alpha = 0}}{\text{.8}}$
$\therefore {\text{i = 1 + (2}} - 1{\text{)0}}{\text{.8}}$ (n = 2, due to the dissociation of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}} \to {\text{N}}{{\text{H}}_3} + {\text{HCl}}$ )
$\Rightarrow$ ${\text{i = 1 + (1)0}}{\text{.8}}$
$\Rightarrow$i = 1.8
Now, we are given 1 litre of 0.1 N HCl solvent, we can calculate the mass of
Normality = M (molarity) $\times$ valence factor
$\Rightarrow$M = $\dfrac{{{\text{Normaity}}}}{{{\text{valence factor }}}}$
$\Rightarrow$M = 0.1 (valence factor of HCl is 1)
Now we have to calculate the mass of HCl
$\Rightarrow$M(Molarity) = $\dfrac{{{\text{Given mass}}}}{{{\text{molar mass }}}}$ (since we have 1 litre of the solution we have ignored the value 1 in the denominator)
$\Rightarrow$Given mass = molar mass $\times$ Molarity
$\Rightarrow$Given mass = 36.46 $\times$ 0.1
$\Rightarrow$Given mass = 3.646
Now, we can calculate the value of molality as we are given the 0.025 mol of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$ solute
$\Rightarrow {\text{m}} = {\text{ }}\dfrac{{{\text{0}}{\text{.025}}}}{{3.646}}{\text{ }}$
$\Rightarrow$ m = 0.00685
Substituting all the values in the formula for ${{\Delta T}}$ we get;
$\Rightarrow$ ${{\Delta }}{{\text{T}}_{\text{f}}}{\text{ = i}}{{\text{K}}_{\text{f}}}{\text{m}}$
$\Rightarrow$ ${{\Delta }}{{\text{T}}_{\text{f}}}$ = ${\text{ 1}}{\text{.8}} \times {\text{1}}{\text{.85}} \times 0.00685$
$\Rightarrow$ ${{\Delta }}{{\text{T}}_{\text{f}}}$ = 0.2281 (we will write only to the second decimal by rounding off the obtained value)
$\Rightarrow$ ${{\Delta }}{{\text{T}}_{\text{f}}}$ = 0.23$^ \circ {\text{C}}$
this value is of difference in freezing temperature, to get the freezing point of the solution we have to subtract the obtained value by ${0^ \circ }{\text{C}}$(freezing point of water)
$\Rightarrow$ ${{\Delta }}{{\text{T}}_{\text{f}}}$= ${\text{T}}_{\text{f}}^ \circ - {{\text{T}}_{\text{f}}}$
$\Rightarrow$ ${{\text{T}}_{\text{f}}}$ = ${\text{T}}_{\text{f}}^ \circ -$ ${{\Delta T}}$
$\Rightarrow$ ${{\text{T}}_{\text{f}}}$ = ${0^ \circ }{\text{C}}$ $-$ 0.23$^ \circ {\text{C}}$
$\Rightarrow$ ${{\text{T}}_{\text{f}}}$ = $-$ 0.23$^ \circ {\text{C}}$

Hence, the correct answer is option (C) i.e., $-$ 0.23$^ \circ {\text{C}}$

Note: Many times we stop our calculation after obtaining the value of ${{\Delta }}{{\text{T}}_{\text{f}}}$ by assuming it to be the final answer, but it is not the final answer unless the question asks to calculate the difference in freezing point i.e, the value of ${{\Delta }}{{\text{T}}_{\text{f}}}$. And if the question asks for the value of freezing point after the adding of solute (i.e., the value of ${{\text{T}}_{\text{f}}}$) like in our question we have to calculate one more step by using the other formula of ${{\Delta }}{{\text{T}}_{\text{f}}}$= ${\text{T}}_{\text{f}}^ \circ - {{\text{T}}_{\text{f}}}$