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NEET CHEMISTRYEasy

The molar conductance of a solution, given its conductivity (0.248 S m–10.248\text{ S m}^{–1}) and concentration (0.2 mol m–30.2\text{ mol m}^{–3}) is:

A

0.124 S cm2 mol–10.124\text{ S cm}^2\text{ mol}^{–1}

B

1.24 S m2 mol–11.24\text{ S m}^2\text{ mol}^{–1}

C

124 S cm2 mol–1124\text{ S cm}^2\text{ mol}^{–1}

D

124 S m2 mol–1124\text{ S m}^2\text{ mol}^{–1}

Step-by-Step Solution

Given: Conductivity, κ=0.248 S m−1\kappa = 0.248\text{ S m}^{-1} Concentration, c=0.2 mol m−3c = 0.2\text{ mol m}^{-3}

Molar conductivity (Λm\Lambda_m) is given by the formula: Λm=κc\Lambda_m = \frac{\kappa}{c}

Substituting the given values: Λm=0.248 S m−10.2 mol m−3\Lambda_m = \frac{0.248\text{ S m}^{-1}}{0.2\text{ mol m}^{-3}} Λm=1.24 S m2 mol−1\Lambda_m = 1.24\text{ S m}^2\text{ mol}^{-1}

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