Question

A potential difference of 20 V applied to the ends of a column of $\frac{M}{10}\text{AgN}{\text{O}}_3$ solution, 4 cm in diameter and 12 cm in length gave a current of 20 ampere. What is the molar conductance $\left(c{m}^2{\Omega }^{-1}mo{l}^{-1}\right)$ of the solution?

• A. 9554
• B. 105.5
• C. 115.5
• D. 125.5

Answer 1

Answer: (A)

9554

By ohm's law, $V=IR$

$R=\frac{20}{20}=1\Omega$

Area of section $=\pi r^2=3.14\times 2\times 2=12.56$ $cm^2

Length of cell $12$ $cm$

Conductivity $=\frac{1}{R}\times \frac{l}{A}=\frac{1}{1}\times \frac{12}{12.56}=0.9554$ $Sc{m}^{-1}$

Molar Conductivity $=\frac{K}{C}\times 1000=\frac{0.9554}{0.1}\times 1000=9554$ $c{m}^2\Omega mo{l}^{-1}$