Question

To $500c{m}^3$ of water $3.0\times {10}^{-3}kg$ of acetic acid is added. If $23\%$ of acetic acid is dissociated, what will be the depression of freezing point?

[$K_f$ and density of water are $1.86Kkgmo{l}^{-1}$ and $0.997gc{m}^{-3}$, respectively]

• A. $0.23K$
• B. $0.33K$
  
• C. $0.98K$
• D. $0.68K$

Answer 1

Answer: (A)

$0.23K$

Molar mass of acetic acid is $60g/mol$.

Number of moles of acetic acid $=\frac{mass}{molarmass}=\frac{3}{60}=0.05$ moles.

Density of water is $0.997$ g/ml. $500$ ml of water corresponds to $500\times 0.997=498.5g$ or $0.4985$ kg

Molality is the number of moles of acetic acid divided by the mass of water (in kg).

Molality $m=\frac{0.05}{0.4985}=0.1m$

$23$ % dissociation of acetic acid means degree of dissociation $\alpha =0.23$

The vant Hoff factor $i=[1+(n-1\left)\alpha \right]=[1+(2-1)\times 0.23]=1.23$

The depression in the freezing point $\Delta T_f=i\times k_f\times m=1.23\times 1.86\times 0.1=0.23K$

Hence, the correct option is $A$