Question

The conductivities at infinite dilution of $N{H}_{4}Cl,NaOH$ and $NaCl$ are $130,218,120$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$. If equivalent conductance of $\frac{N}{100}$ solution of $N{H}_{4}OH$ is $10$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$, then degree of dissociation of $N{H}_{4}OH$ at this dilution is:

• A. $0.005$
• B. $0.043$
• C. $0.01$
• D. $0.02$

Answer 1

The correct option is B $0.043$

${\lambda }_m^o\left(N{H}_{4}OH\right)={\lambda }_m^o(N{H}_4^{+}+{\lambda }_m^o{)}^{-}$

Now

${\lambda }_m^o\left(N{H}_{4}OH\right)={\lambda }_m^o\left(N{H}_{4}Cl\right)+{\lambda }_m^o\left(NaOH\right)$

$-{\lambda }_m^o\left(Nacl\right)$

${\lambda }_m^o\left(N{H}_{4}OH\right)=130+218-120$

${\lambda }_m^o\left(N{H}_{4}OH\right)=228$

Equivalent conductance $={\lambda }_m$

$\Rightarrow 10oh{m}^{-1}c{m}^{2}e{q}^{-1}$

Degree of dissoocition $=\frac{\lambda m}{{\lambda }_m^o}$

$\Rightarrow \frac{10}{228}$

$\Rightarrow 0.043$