Question

A solution containing $0.85\, g$ of $Z n C l_{2}$ in $125.0 \,g$ of water freezes at- $0.23^{\circ} C$. The apparent degree of dissociation of the salt is $\left(K_{f}\right.$ for water $=1.86 \,K\,kg \,mol ^{-1}$, atomic mass: $Zn =65.3$ and $Cl =35.5$ )

• A. 1.36%
• B. 73.5%
• C. 7.35%
• D. 2.47%

Answer 1

Answer: (B)

Mol. $w t=\frac{k_{f} \times w \times 1000}{\Delta T_{f} \times W}$
$=\frac{1.86 \times 0.85 \times 1000}{0.23 \times 125} \approx 55\, gm$
Where
$w=0.85\, g$
$W=125 \,g$
$\Delta T_{f}=0^{\circ} C-\left(-23^{\circ} C\right)=23^{\circ} C$
Now, $i=\frac{M_{\text {normal }}}{M_{\text {observed }}}=\frac{136.3}{55}=2.47$

Van't Hoff factor (i)
$=\frac{1-\alpha+\alpha+2 \alpha}{1}=2.47 $
$\therefore \alpha=0.735=73.5 \%$