Question

The electrical resistance of a column of $0.05$M NaOH solution of diameter $1$cm and length $50$ cm is $5.55\times {10}^3\Omega$. Its molar conductivity will be:

• A. $329.6Sc{m}^{2}mo{l}^{-1}$
• B. $229.6Sc{m}^{2}mo{l}^{-1}$
• C. $429.6Sc{m}^{2}mo{l}^{-1}$
• D. $529.6Sc{m}^{2}mo{l}^{-1}$

Answer 1

Answer: (B)

$229.6Sc{m}^{2}mo{l}^{-1}$

Area=$\pi r^2$

=$3.14\times {\left(0.5\right)}^2=0.785c{m}^2$

=$50$

$R=\frac{\rho l}{A}$

Implies that

$\rho =\frac{RA}{l}$=$\frac{5.55\times {10}^3\Omega \times 0.785c{m}^2}{50cm}$=$87.135\Omega cm$

Conductivity $K=\frac{1}{\rho }=\left(\frac{1}{87.135}\right)Sc{m}^{-1}$$0.11845c{m}^{-1}$

Molarconductivuty ${\wedge }_m=\frac{k\times 1000}{c}c{m}^3{L}^{-1}$

Implies that

=$\frac{0.01148Sc{m}^{-1}\times 1000c{m}^3{L}^{-1}}{0.05mol{L}^{-1}}$

Hence

${\wedge }_m=229.6Sc{m}^{2}mo{l}^{-1}$