Question

Molar conductivity of a solution $1.26\times {10}^2{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$ , Its molarity is 0.01. its specific conductivity will be :

• A. $1.26\times {10}^{-5}$
• B. $1.26\times {10}^{-3}$
• C. $1.26\times {10}^{-4}$
• D. $6.3\times {10}^{-3}$

Answer 1

Answer: (B)

$1.26\times {10}^{-3}$

The relationship between the molar conductivity and the specific conductance is ${\Lambda }_m=\frac{\kappa \times 1000}{M}$.

Substitute ${\Lambda }_m=1.26\times {10}^2{\Omega }^{-1}{cm}^2{mol}^{-1}$ and $M=0.01M$ in the above equation:

$1.26\times {10}^2=\frac{\kappa \times 1000}{0.01}$

$\kappa =1.26\times {10}^{-3}$.