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Resistance of  0.2  M  solution of an electrolyte is  50 \Omega.   The specific conductance of the solution is 1.4 S m-1.  The resistance of 0.5 M solution of the same electrolyte is 280 \Omega. The molar conductivity of 0.5 M solution of the electrolyte in  S m2 mol-1 is :

Option: 1

5 x 10-4

Option: 2

5 x 10-3

Option: 3

5 x 103

Option: 4

5 x 102

Answers (1)


Specific conductance,

\sigma =1.4Sm^{-1}=1.4\times 10^{-2}\: 5cm^{-1}


\rho =\frac{1}{\kappa}=\frac{1}{1.4\times 10^{-2}}\ \Omega cm

Resistance, R=\frac{\rho l}{A}

Now, for a 0.5 M solution, R=280 \Omega

\kappa =\frac{1}{\rho}=\frac{1}{R}\times \frac{l}{A}=\frac{1}{280}\times 50\times 1.4\times 10^{-2}

=2.5\times 10^{-3}Scm^{-1}

\therefore molar conductivity,

\mu=\frac{\kappa\times 1000}{c}=\frac{2.5\times 10^{-3}\times 1000}{0.5}

\mathrm{5\ Scm^{2}mol^{-1}}

\mathrm{=5\times 10^{-4} Sm^{2} mol^{-1}}

Therefore, the correct option is (1).

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