Question

What is the freezing point of $0.4$ molal solution of acetic acid of benzene in which it dimerise to the extent of $80$%? Freezing point of benzene is $278.4K$ and its molar heat of fusion is $10.042kJ$ ${mol}^{-1}$.

• A. 277.25 K
• B. 282.25 K
• C. 180.6 K
• D. 152.25 K

Answer 1

The correct option is A 277.25 K

$K_f=\frac{R{T}_f^{2}M}{1000\times \Delta H_{fusion}}$

Here, $K_f=\frac{8.314\times (278.4{)}^2\times 78}{1000\times 10.042\times {10}^3}$

$=5KKg/mol$

$2C{H}_{3}COOH⟶(C{H}_{3}COOH{)}_2$

$\Rightarrow \alpha =\frac{i-1}{\frac{1}{n}-1}=\frac{i-1}{1/2-1}=0.8$

$\Rightarrow i-1=-0.4$

$\Rightarrow i=0.6$

$\Delta T_f=K_f\times m\times i$

$=5\times 0.6\times 0.4$

$=1.2$

$\therefore T_f^1=278.4-1.2=277.2K$