Question

Resistance of $0.2 \, M$ solution of an electrolyte is $50 \Omega$. The specific conductance of the solution is $1.3\, S m^{-1}$. If resistance of the $0.4\, M$ solution of the same electrolyte is $260\, \Omega$ , its molar conductivity is

• A. $6250\, S\, m^{2}\, mol^{-1}$
• B. $6.25 \times 10^{-4} S\, m^{2} \,mol^{-1}$
• C. $625 \times 10^{-4} S\, m^{2} \,mol^{-1}$
• D. $62.5\, S \,m^{2}\, mol^{-1}$

Answer 1

Answer: (B)

For $0.2 M$ solution,
Specific conductance (k ) = Conductance $\left(\frac{1}{R}\right)\times$ Cell constant$\left(l a\right) $
$\Rightarrow K \frac{1}{50} \times\frac{l}{a}=1.3\, S\, m^{-1} $
$\Rightarrow \frac{1}{a}=1.3\times50\, m^{-1}$
$\frac{1}{a}=1.3 \times50\times10^{-2} cm^{-1}$
for $0.4$ solution $R=260 \Omega$
$R=\rho \frac{1}{a}$
$\Rightarrow \frac{1}{\rho}=k=\frac{1}{R} \times\frac{1}{a}$
$k=\frac{1}{260} \times1.3\times50\times10^{-2}S Cm^{-1}$
Molar conductivity of the solution is given by
$\wedge_{m} =\frac{k \times1000}{M}$
$=\frac{1}{260}\times\frac{1.3\times50\times10^{-2}\times1000}{4}$
$=6.25\, S\, cm^{2} mol^{-1}$
$=6.25 \times10^{-4}S\, m^{2} mol^{-1}$