Question

The molar conductivity of $0.05M$ $BaC{l}_2$ solution at ${25}^{o}C$ is $223{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$.

What is its conductivity ?

• A. $0.01115{\Omega }^{-1}c{m}^{-1}$
• B. $0.02225{\Omega }^{-1}c{m}^{-1}$
• C. $0.03335{\Omega }^{-1}c{m}^{-1}$
• D. $0.04445{\Omega }^{-1}c{m}^{-1}$

Answer 1

The correct option is A $0.01115{\Omega }^{-1}c{m}^{-1}$

${\Lambda }_m$ (Molar Conductivity) $=223{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$

$C$ (Concentration) $=0.05M$

$k$ (Conductivity) $=?$

${\Lambda }_m=\frac{1000\times k}{C}$

$223=\frac{1000\times k}{0.05}$

$\frac{223\times 0.05}{1000}=k$

$\frac{11.15}{1000}=k$

$0.01115{\Omega }^{-1}c{m}^{-1}=k$

The conductivity is $0.01115{\Omega }^{-1}c{m}^{-1}$