Question

The conductivity of $0.01{\text{mol/dm}}^3$ aqueous acetic acid at $300K\text{is}19.5\times {10}^{-5}{\text{ohm}}^{-1}{\text{cm}}^{-1}$ and the maximum limiting molar conductivity of acetic acid at the same temperature is $390{\text{ohm}}^{-1}{\text{cm}}^2{\text{mol}}^{-1}$. The degree of dissociation of acetic acid is:

• A. $0.5$
• B. $0.05$
• C. $5\times {10}^{-3}$
• D. $5\times {10}^{-7}$

Answer 1

The correct option is B $0.05$

Concentration $\left(C\right)=0.01{\text{mol dm}}^{–3}=10{\text{mol m}}^{–3}$
Conductivity $\left(K\right)=19.5\times {10}^{-5}{\text{S cm}}^{-1}$
$=19.5\times {10}^{-3}{\text{S m}}^{-1}$
Limiting molar conductivity $\left({\Lambda }_m^{○}\right)=390{\text{S cm}}^2{\text{mol}}^{-1}$
$=390\times {10}^{–4}{\text{S m}}^2{\text{mol}}^{–1}$
Molar conductivity ${\Lambda }_m=\frac{K}{C}=\frac{19.5\times {10}^{-3}}{10}$
$=19.5\times {10}^{–4}{\text{S m}}^2{\text{mol}}^{–1}$
Degree of dissociation$\left(\alpha \right)\frac{{\Lambda }_m}{{\Lambda }_m^{○}}=\frac{19.5\times {10}^{-4}}{390\times {10}^{-4}}$
$\alpha =0.05$