JEE Main & Advanced Sample Paper JEE Main - Mock Test - 15

The equivalent conductivity of \[N/10\]solution of acetic acid at \[25{}^\circ C\]is\[14.3\text{ }oh{{m}^{-1}}\text{ }c{{m}^{2}}\text{ }e{{q}^{-1}}\]. What will be the degree of dissociation of acetic acid? \[({{\wedge }^{\infty }}_{C{{H}_{3}}COOH}=390.71\,oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}})\]

A) \[3.66%\]

B) \[3.9%\]

C) \[2.12%\]

D) \[0.008%\]

Correct Answer: A

Solution :

\[{{\wedge }^{\infty }}_{C{{H}_{3}}COOH}=390.71\,oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}\]

\[{{\wedge }^{\infty }}_{C{{H}_{3}}COOH}=14.3\,oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}\]

Degree of dissociation \[(\alpha )=\frac{{{\wedge }^{c}}}{{{\wedge }^{\infty }}}\]

\[=\frac{14.3}{390.71}=0.0366\] i.e. \[3.66%\]