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Chemistry Question on Conductance

The conductivity of 0.20 M KCl0.20 \,M\, K C l solution at 298 K298\, K is 0.0248 S cm−10.0248\, S\, cm ^{-1}. What will be its molar conductivity?

A

124 S cm2124 \,S\,cm ^{2}

B

124 cm−1124\, cm ^{-1}

C

124 ohm−1 cm2124\, ohm ^{-1} \,cm ^{2} equiv −1^{-1}

D

124 S cm2 mol−1124\, S\,cm ^{2}\, mol ^{-1}

Answer

124 S cm2 mol−1124\, S\,cm ^{2}\, mol ^{-1}

Explanation

Solution

Molar conductivity, Λm=k×1000C\Lambda_{m}=\frac{k \times 1000}{C} Given, k=0.0248 S cm−1k=0.0248\, S\, cm ^{-1} C=0.20 MC=0.20 \,M ∴Λm=0.0248×10000.20\therefore \Lambda_{m}=\frac{0.0248 \times 1000}{0.20} =124 S cm2 mol−1=124 \,S\,cm ^{2} \,mol ^{-1}