Question

In a conductivity cell the two platinum electrodes, each of area 10 sq. cm. are fixed 1.5 cm apart. The cell contained 0.05 N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of 50 ohms, the equivalent conductance of the salt solution will be:

• A. $120mhoc{m}^{2}e{q}^{-1}$
• B. $100mhoc{m}^{2}e{q}^{-1}$
• C. $220mhoc{m}^{2}e{q}^{-1}$
• D. none of these

Answer 1

Answer: (A)

$120mhoc{m}^{2}e{q}^{-1}$

$0.05$ N solution means $0.05$ g eq are present in $1000$ mL.
Hence, 1 g eq will be present in $V=\frac{1000}{0.05}=20000mL$.
The specific conductivity $\kappa =\frac{l}{a\times R}=\frac{1.5}{5\times 50}=6\times {10}^{-3}mhoc{m}^{-1}$
The equivalent conductivity is $\Lambda =\kappa \times V=6\times {10}^{-3}mhoc{m}^{-1}\times 20000c{m}^3/eq=120mhoc{m}^2/eq$.