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Chemistry Question on Electrochemistry

Resistance of 0.2 M0.2\, M solution of an electrolyte is 50 Ω50 \,\Omega. The specific conductance of the solution of 0.5 M0.5 \,M solution of same electrolyte is 1.4 S m−11.4\, S \, m^{-1} and resistance of same solution of the same electrolyte is 280 Ω280 \,\Omega . The molar conductivity of 0.5 M0.5\, M solution of the electrolyte in S m2 mol−1S\,m^2 \,mol^{-1} is

A

5×10−45 \times10^{-4}

B

5×10−35 \times10^{-3}

C

5×1035 \times10^{3}

D

5×1025 \times10^{2}

Answer

5×10−45 \times10^{-4}

Explanation

Solution

For 0.2 M0.2 \,M solution

R=50 ΩR =50\, \Omega
σ=1.4 S m−1=1.4×10−2S cm−1\sigma=1.4 \,S \,m ^{-1}=1.4 \times 10^{-2} S \,cm ^{-1}
⇒ρ=1σ=11.4×10−2Ω cm\Rightarrow \rho=\frac{1}{\sigma}=\frac{1}{1.4 \times 10^{-2}} \Omega \,cm

Now, R=ρlaR =\rho \frac{l}{ a }
⇒la=Rρ=50×1.4×10−2\Rightarrow \frac{l}{ a }=\frac{ R }{\rho}=50 \times 1.4 \times 10^{-2}

For 0.50.5 M solution

R=280 ΩR =280 \,\Omega
σ=?\sigma=?
la=50×1.4×10−2\frac{l}{ a }=50 \times 1.4 \times 10^{-2}
⇒R=ρla\Rightarrow R =\rho \frac{l}{ a }
⇒1ρ=1R×la\Rightarrow \frac{1}{\rho}=\frac{1}{ R } \times \frac{l}{ a }
⇒σ=1280×50×1.4×10−2\Rightarrow \sigma=\frac{1}{280} \times 50 \times 1.4 \times 10^{-2}
=1280×70×10−2=\frac{1}{280} \times 70 \times 10^{-2}
=2.5×10−3 S cm−1=2.5 \times 10^{-3} \,S \,cm ^{-1}

Now , λm=σ×1000M\lambda_{ m }=\frac{\sigma \times 1000}{ M }
=2.5×10−3×10000.5=\frac{2.5 \times 10^{-3} \times 1000}{0.5}
=5 S cm2 mol−1=5\, S \,cm ^{2} \,mol ^{-1}
=5×10−4 S m2 mol−1=5 \times 10^{-4} \,S \,m ^{2} \,mol ^{-1}