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Chemistry Question on Thermodynamics

The equivalent conductivity of a solution containing 2.54 g2.54\, g of CuSO4CuSO _{4} per LL is 91.0 W−1cm2eq−191.0\, W ^{-1} cm ^{2} eq ^{-1}. Its conductivity would be

A

2.9×10−3 Ω−1 cm−12.9 \times 10^{-3} \, \Omega^{-1} \, cm^{-1}

B

1.8×10−2 Ω−1 cm−11.8 \times 10^{-2} \, \Omega^{-1} \, cm^{-1}

C

2.4×10−4 Ω−1 cm−12.4 \times 10^{-4} \, \Omega^{-1} \, cm^{-1}

D

3.6×10−3 Ω−1 cm−13.6 \times 10^{-3} \, \Omega^{-1} \, cm^{-1}

Answer

2.9×10−3 Ω−1 cm−12.9 \times 10^{-3} \, \Omega^{-1} \, cm^{-1}

Explanation

Solution

We know that, κ=Λeq.C\kappa = \Lambda_{eq} . C Given, Λeq=91.0 Ω−1 cm2 eq−1\Lambda eq = 91.0 \, \Omega^{-1} \, cm^2 \, eq^{-1} κ=(91Ω−1cm2eq−1)\kappa = \left( 91 \Omega^{-1} cm^2 eq^{-1}\right) (2.541592×1000ecm−3)\left( \frac{2.54}{ 159 2 \times1000} ecm^{-3}\right) =2.9×10−3Ω−1cm−1 = 2.9 \times10^{-3} \Omega^{-1} cm^{-1}