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Chemistry Question on Conductance

The molar conductivities of NaOH,NaCl {NaOH, NaCl} and BaCl2 { BaCl2} at infinite dilution are 2.481×10−2 S m2 mol−1,1.265×10−2 S m2 mol−1 {2.481 \times 10^{-2}\, S \, m^2\, mol^{-1}, 1.265 \times 10^{-2} \, S\, m^2\, mol^{-1}} and 2.800×10−2 S m2 mol−1 {2.800 \times 10^{-2} \, S \, m^2\, mol^{-1}} respectively. The molar conductivity of Ba(OH)2 {Ba(OH)_2} at infinite dilution will be

A

\ce5.232×10−2 S m2 mol−1\ce{5.232 \times 10^{-2} \, S \, m^2\, mol^{-1}}

B

\ce9.654×10−2 S m2 mol−1\ce{9.654 \times 10^{-2} \, S \, m^2\, mol^{-1}}

C

\ce4.016×10−2 S m2 mol−1\ce{4.016 \times 10^{-2} \, S \, m^2\, mol^{-1}}

D

\ce1.145×10−2 S m2 mol−1\ce{1.145 \times 10^{-2} \, S \, m^2\, mol^{-1}}

Answer

\ce5.232×10−2 S m2 mol−1\ce{5.232 \times 10^{-2} \, S \, m^2\, mol^{-1}}

Explanation

Solution

⋀m∘[Ba(OH)2]\bigwedge^\circ_m { [Ba(OH)2]}
=⋀m∘(BaCl2)+2⋀m∘(NaOH)−2⋀m∘(NaCl)= \bigwedge^\circ_m { (BaCl_2)} + 2\bigwedge^\circ_m { (NaOH)} - 2\bigwedge^\circ_m {(NaCl)}
= (2.8+2×2.481−2×1.265)×10−2 S m2 mo1−1 { (2.8 + 2 \times 2.481 - 2 \times 1.265) \times 10^{-2}\, S \, m^2\, mo1^{-1}}
= (2.8+4.962−2.53)×10−2 S m2 mol−1 { (2.8 + 4. 962 - 2.53) \times 10^{-2} \, S\, m^2 \, mol^{-1}}
= 5.232×10−2 S m2 mol−1 {5.232 \times 10^{-2} \, S\, m^2 \, mol^{-1}}