Question

In a 0.2 molal aqueous solution of a weak acid HX, the degree of ionization is 0.3 at room temperature. Taking $K_f$ for water as $1.85{}^{o}C{m}^{-1}$, the freezing point of the solution will be nearest to

• A. $-0.481{}^{\circ }C$
• B. $-0.360{}^{\circ }C$
• C. $-0.260{}^{\circ }C$
• D. $-0.150{}^{\circ }C$

Answer 1

The correct option is A $-0.481{}^{\circ }C$

We know,
$\Delta T_f=i{K}_{f}m$
Again, we know, $i=1+(n-1)\alpha$
Where, $\alpha$ = degree of ionisation
So here $i=1+(2-1)0.3=1.3$
Again, since the solution is 0.2 molal
Thus,
$\Delta T_f=1.3\times 1.85\times 0.2=0.481{}^{o}C$
Again,
$\Delta T_f=T_{f\left(solvent\right)}-T_{f\left(solution\right)}$
$\Rightarrow 0.481=0-T_{f\left(solution\right)}$
$\Rightarrow T_{f\left(solution\right)}=-0.481{}^{o}C$