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Chemistry Question on Conductance

Specific conductance of 0.1 N KCl0.1 \,N \,KCl solution at 23∘C23^{\circ} C is 0.012 ohm−1 cm−10.012 \,ohm ^{-1} \,cm ^{-1}. Resistance of cell containing the solution at same temperature was found to be 55 ohm55 \,ohm. The cell constant is:

A

0.0616 cm−10.0616\, cm ^{-1}

B

0.616 cm−10.616 \,cm ^{-1}

C

6.16 cm−16.16 \,cm ^{-1}

D

616 cm−1616 \,cm ^{-1}

Answer

0.616 cm−10.616 \,cm ^{-1}

Explanation

Solution

κ=GG∗\kappa= GG ^{*} where κ=\kappa= specific conductance =0.0112 ohm−1 cm−1=0.0112\, ohm ^{-1} \,cm ^{-1} G=G = conductance =155=0.01818 ohm−1=\frac{1}{55}=0.01818 \,ohm ^{-1} G∗=G ^{*}= cell constant 0.0112=0.01818×G∗0.0112=0.01818 \times G ^{*} G∗=0.616 cm−1G ^{*}=0.616\, cm ^{-1}