Question

Given $l/a = 0.5 cm^{-1} ,R = 50$ ohm, $N = 1.0$. The equivalent conductance of the electrolytic cell is

• A. $100$ ohm$^{-1} cm^{2} \,g \,Eq^{-1}$
• B. $20$ ohm$^{-1} cm^{2} \,g \,Eq^{-1}$
• C. $300$ ohm$^{-1} cm^{2}\, g \, Eq^{-1}$
• D. $10$ ohm$^{-1} cm^{2}\, g\,Eq^{-1}$

Answer 1

Answer: (D)

$l/a =0.5 \,cm^{-1}, R=50\Omega$
$\rho =\frac{Ra}{l} =\frac{50}{0.5}=100$
$\Lambda =k \times \frac{1000}{N} =\frac{1}{\rho} \times \frac{1000}{N} =\frac{1}{100} \times \frac{1000}{1}$
$=10 \,ohm^{-1} Cm^{2} \, g\,Eq^{-1}$