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Physics Question on Electric Current

Specific conductance of 0.1 M NaCl0.1\, M\, NaCl solution is 1.01×10−2 ohm−1 cm−11.01 \times 10^{-2}\, ohm^{-1}\, cm^{-1}. Its molar conductance in ohm−1 cm2 mol−1ohm^{-1}\, cm^2 \,mol^{-1} is

A

1.01×1021.01\times10^{2}

B

1.01×1031.01 \times 10^{3}

C

1.01×1041.01 \times 10^{4}

D

1.011.01

Answer

1.01×1021.01\times10^{2}

Explanation

Solution

Λm=κ×1000M\Lambda_{m}=\frac{\kappa\times1000}{M} =1.01×10−2×10000.1=\frac{1.01\times10^{-2}\times1000}{0.1} =1.01×102 ohm−1 cm2 mol−1=1.01\times10^{2}\,ohm^{-1}\,cm^{2}\,mol^{-1}