Question

The freezing point of a solution containing $0.2$ g of acetic acid in $20.0$ g benzene is lowered by ${0.45}^{o}C$. The degree of association of acetic acid in benzene is:

[Assume acetic acid dimerizes in benzene and $K_f$ for benzene $=5.12$ K kg mol${}^{-1}$]

• A. $94.5$%
• B. $54.9$%
• C. $78.2$%
• D. $100$%

Answer 1

The correct option is A $94.5$%

The molar mass of acetic acid is $60g/mol$.
The molality of the solution is the number of moles of solute divided by the mass of solvent (in kg).
The molality of the solution will be $\frac{0.2\times 1000}{60\times 20.0}$
$\Delta T_f=i{K}_{f}m$

$0.45=i\times 5.12\times \frac{0.2\times 1000}{60\times 20.0}$

$i=0.5275$
The degree of association of acetic acid in benzene is:

$\alpha =\frac{i-1}{\frac{1}{n}-1}$
Here, n is the number of molecules of acetic acid that associate.
$\alpha =\frac{0.5275-1}{\frac{1}{2}-1}=0.945$
Hence, the degree of association of acetic acid in benzene is $94.5$%.