Question
Resistance of a conductivity cell filled with a solution of an electrolyte of concentration $$0.1 M$$ is $$100\,\Omega .$$ The conductivity of this solution is $$1.29\,S\,{m^{ - 1}}.$$ Resistance of the same cell when filled with $$0.2 M$$ of the same solution is $$520\,\Omega .$$ The
molar conductivity of $$0.2 M$$ solution of electrolyte will be
A.
$$1.24 \times {10^{ - 4}}S\,{m^2}\,mo{l^{ - 1}}$$
B.
$$12.4 \times {10^{ - 4}}S\,{m^2}\,mo{l^{ - 1}}$$
C.
$$124 \times {10^{ - 4}}S\,{m^2}\,mo{l^{ - 1}}$$
D.
$$1240 \times {10^{ - 4}}S\,{m^2}\,mo{l^{ - 1}}$$
Answer :
$$12.4 \times {10^{ - 4}}S\,{m^2}\,mo{l^{ - 1}}$$
Solution :
$$R = 100\,\Omega ,\,\kappa = \frac{1}{R}\left( {\frac{1}{a}} \right),$$ $$\frac{1}{a}\left( {{\text{cell}}\,{\text{constant}}} \right) = 1.29 \times 100{m^{ - 1}}$$
$${\text{Given,}}\,R = 520\,\Omega ,\,C = 0.2M,$$ $$\mu {\text{(molar conductivity) = ?}}$$
$$\mu {\text{ = }}\kappa \times {\text{V}}$$ ( $$\kappa $$ can be calculated as $$\kappa = \frac{1}{R}\left( {\frac{1}{a}} \right)$$ now cell constant is known. )
$$\eqalign{
& {\text{Hence,}} \cr
& \mu = \frac{1}{{520}} \times 129 \times \frac{{1000}}{{0.2}} \times {10^{ - 6}}{m^3} \cr
& \,\,\,\,\, = 12.4 \times {10^{ - 4}}\,S{m^2}mo{l^{ - 1}} \cr} $$