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Q. The conductivities at infinite dilution of $NH _{4} Cl , NaOH$ and $NaCl$ are $130,218,120 \,ohm ^{-1} \,cm ^{2} eq ^{-1} .$ If equivalent conductance of $\frac{ N }{100}$ solution of $NH _{4} OH$ is $10$ , then degree of dissociation of $NH _{4} OH$ at this dilution is

Electrochemistry

Solution:

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$130+218=120+x$
$\alpha=\frac{\wedge_{ m }^{ C }}{\wedge_{ m }^{\infty}}=\frac{10}{228}=0.043$
$x =228$ of $\wedge_{ NH _{4} OH }^{\infty}$