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Physics Question on Current electricity

The resistance of 0.01 N0.01\, N solution of an electrolyte was found to be 220 ohm220\, ohm at 298 K298\, K using a conductivity cell with a cell constant of 0.88 cm−10.88\,cm^{-1}. The value of equivalent conductance of solution is -

A

400 mho cm2 g eq−1400\, mho\, cm^2 \,g \,eq^{-1}

B

295 mho cm2 g eq−1295\, mho\, cm^2 \,g \,eq^{-1}

C

419 mho cm2 g eq−1419\, mho\, cm^2 \,g \,eq^{-1}

D

425 mho cm2 g eq−1425\, mho\, cm^2 \,g \,eq^{-1}

Answer

400 mho cm2 g eq−1400\, mho\, cm^2 \,g \,eq^{-1}

Explanation

Solution

Λeq=k×1000N=1R×la×1000N\Lambda_{eq} = k \times \frac{1000}{N} = \frac{1}{R}\times \frac{l}{a}\times \frac{1000}{N} =1R×= \frac{1}{R} \times cell constant ×1000N\times \frac{1000}{N} =1220×0.88×10000.01= \frac{1}{220}\times0.88\times \frac{1000}{0.01} 400 mho cm2 g eq−1400\, mho\, cm^{2} \,g \,eq^{-1}