Question

$1.0g$ of a monobasic acid $HA$ in $100g$ water lowers the freezing point by $0.155K$. If $0.75g$, of same acid requires $15$ mL of $N/5NaOH$ solution for complete neutralisation then %, degree of ionization if acid is ($K_{f}of{H}_{2}O=1.86Kkgmo{l}^{-1}$):

• A. 20%
• B. 25%
• C. 40%
• D. 50%

Answer 1

The correct option is B 25%

(b) $\Delta T_f=K_f\times \frac{W_{acid}}{M_{acid}\times W_{H_{2}O}}\times 1000$$\therefore M_{acid}\left(observed\right)=120$
Normal molecular mass of acid can calculate as
milli-equivalents of acid = mili-equivalents of base
(for $HA$ and $NaOH,N=M$)
$\frac{0.75}{M_{acid}}\times 1000=15\times \frac{1}{5}$
$\Rightarrow M_{acid\left(normal\right)}=150$
$\therefore i=\frac{150}{120}\Rightarrow 1.25$
$i=1+a\Rightarrow 0.25$
$\therefore$ %$a=25$