Question
In a 0.2 molal aqueous solution of a weak acid $$HX$$ the degree
of ionization is 0.3. Taking $${k_f}$$ for water as 1.85, the freezing
point of the solution will be nearest to
A.
$$ - {\text{ }}{0.360^ \circ }C$$
B.
$$ - {\text{ }}{0.260^ \circ }C$$
C.
$$ + {\text{ }}{0.480^ \circ }C$$
D.
$$ - {\text{ }}{0.480^ \circ }C$$
Answer :
$$ - {\text{ }}{0.480^ \circ }C$$
Solution :
$$\eqalign{
& \Delta {T_f} = {K_f} \times m \times i\,; \cr
& \Delta {T_f} = 1.85 \times 0.2 \times 1.3 = {0.480^ \circ }C \cr
& \therefore \,\,{T_f} = 0 - {0.480^ \circ }C = - {0.480^ \circ }C \cr
& \left( {\mathop {HX}\limits_{1 - 0.3} \rightleftharpoons \mathop H\limits_{0.3}^ + + \mathop {{X^ - }}\limits_{0.3} ,i = 1.3} \right) \cr} $$