Question

The resistance of $0.01\, N$ solution of an electrolyte was found to be $220\, ohm$ at $298\, K$ using a conductivity cell with a cell constant of $0.88\,cm^{-1}$. The value of equivalent conductance of solution is -

• A. $400\, mho\, cm^2 \,g \,eq^{-1}$
• B. $295\, mho\, cm^2 \,g \,eq^{-1}$
• C. $419\, mho\, cm^2 \,g \,eq^{-1}$
• D. $425\, mho\, cm^2 \,g \,eq^{-1}$

Answer 1

Answer: (A)

$\Lambda_{eq} = k \times \frac{1000}{N} = \frac{1}{R}\times \frac{l}{a}\times \frac{1000}{N}$
$= \frac{1}{R} \times$ cell constant $\times \frac{1000}{N}$
$= \frac{1}{220}\times0.88\times \frac{1000}{0.01}$
$400\, mho\, cm^{2} \,g \,eq^{-1}$