Discussion Forum

 Resistance of 0.2 M solution of an electrolyte is 50Ω. The specific conductance of the solution of 0.5M solution of same electrolyte is 1.4 Sm^-1 and resistance of same solution of the same electrolyte is 280Ω. The molar conductivity of 0.5 M solution  of the electrolyte in Sm² mol^-1 is 

Answer:

Explanation:

 In order to solve the problem, calculate the value of cell constant of the first solution and then use this value of cell constant to calculate the value of k of second solution. Afterwards, finally, calculate molar conductivity using value of k and m.

 For first solution,

  k=1.4 Sm^-1 , R=50Ω, M= 0.2

 Specific conductance (k)=   $\frac{1}{R}\times\frac{1}{A}$

$1.4 Sm^{-1}=\frac{1}{50}\times\frac{1}{A}$

$\frac{1}{A}=50\times 1.4 m^{-1}$

  for second solution. 

          R=280,   $\frac{1}{A}=50\times 1.4 m^{-1}$

  $k=\frac{1}{280}\times50\times 1.4 =\frac{1}{4}$

 Now, molar conductivity

   $\lambda_{m}=\frac{K}{1000\times m}= \frac{1}{2000}=5\times 10^{-4}Sm^{2}mol^{-1}$