Question

The molar conductivity of $0.05 M$ of solution of an electrolyte is $200 \Omega^{-1}\, cm^2\, mol^{-1}$. The resistance offered by a conductivity cell with cell constant $(1/3) \,cm^{-1}$ would be about

• A. 11.11 $\Omega$
• B. 22.22 $\Omega$
• C. 33.33 $\Omega$
• D. 44.44 $\Omega$

Answer 1

Answer: (C)

$k =\wedge_{c}=\left(200\, S\, cm^{2} mol^{-1}\right)\left(0.05\, \times10^{-3} mol\, cm^{-1}\right)$
$=0.01 \,S\, cm^{-1}$
$R=\frac{1}{k}\left(\frac{l}{a}\right) =\frac{1}{\left(0.01 \,S\, cm^{-1}\right)}\left(\frac{1}{3}cm^{-1}\right)=33.33\, \Omega$