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Chemistry Question on Solutions

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Ω\Omega. The conductivity of this solution is 1.29Sm−1.1.29 S m^{-1}.. Resistance of the same cell when filled with 0.2 M of the same solution is 520 Ω\Omega. The molar conductivity of 0.02 M solution of the electrolyte will be:

A

124×10−4Sm2mol−1124\times10^{-4} S m^{2} mol^{-1}

B

1240×10−4Sm2mol−11240\times10^{-4} S m^{2} mol^{-1}

C

1.24×10−4Sm2mol−11.24\times10^{-4} S m^{2} mol^{-1}

D

12.4×10−4Sm2mol−112.4\times10^{-4} S m^{2} mol^{-1}

Answer

12.4×10−4Sm2mol−112.4\times10^{-4} S m^{2} mol^{-1}

Explanation

Solution

R=100ΩR=100 \Omega K=1R(1a)K=\frac{1}{R} \left(\frac{1}{a}\right) 1a(cell constant)=1.29×100m−1\frac{1}{a} \left(cell \,constant\right)=1.29\times100 m^{-1} Given R=520Ω;C=0.2MGiven \, R=520 \Omega ; C=0.2 M μ\mu (molar conductivity) = ? μ=K×V\mu=K\times V (K can be calculated as k=1R(1a)k=\frac{1}{R} \left(\frac{1}{a}\right) now cell constant is known) Hence μ=1520×129×10000.2×10−6m3\mu=\frac{1}{520} \times129 \times\frac{1000}{0.2}\times10^{-6} m^{3} =12.4×10−4Sm2mol−1=12.4\times10^{-4} Sm^{2} mol^{-1}