Question

Molar conductance of $0.1M$ acetic acid is $7{ohm}^{-1}{cm}^2{mol}^{-1}$. If the molar conductance of acetic acid at infinite dilution is $380.8{ohm}^{-1}{cm}^2{mol}^{-1}$, the value of dissociation constant will be:

• A. $2.26\times {10}^{-5}mol{dm}^{-3}$
• B. $1.66\times {10}^{-3}mol{dm}^{-3}$
• C. $1.66\times {10}^{-2}mol{dm}^{-3}$
• D. $3.442\times {10}^{-5}mol{dm}^{-3}$

Answer 1

The correct option is D $3.442\times {10}^{-5}mol{dm}^{-3}$

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_{m^{\infty }}}\frac{\left(molarconductance\right)}{(mol.cond.atinfinitedilution)}$

$\alpha =0.0183$

since $\alpha <5$%

$K={C\alpha }^2$

$K=0.1\times {\left(0.0183\right)}^2$

$K=3.4\times {10}^{-5}mol{dm}^{-2}$