Question

# Limiting molar conductivity for some ions is given(in S cm2 mol−1):Na+=50.1, Cl−=76.3, H+=349.6, CH3COO−=40.9, Ca2+−119.0.What will be the limiting molar conductivities (Λom) of CaCl2,CH3COONa and NaCl respectively?

A
97.65,111.0 and 242.8 S.cm2 mol1
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B
195.3,182.0 and 26.2 S.cm2 mol1
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C
271.6,91.0 and 126.4 S.cm2 mol1
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D
119.0,1024.5 and 9.2 S.cm2 mol1
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Solution

## The correct option is C 271.6,91.0 and 126.4 S.cm2 mol−1We know,According to Kohlraush's law at infinite dilution,Λ0m(CaCl2)=Λ0Ca2++2Λ0Cl−Λ0m(CaCl2)=119+2×76.3=271.6 S cm2 mol−1Also,Λ0m(CH3COONa)=Λ0CH3COO−+Λ0Na+Λ0m(CH3COONa)=40.9+50.1=91 S cm2 mol−1Similarly,Λ0m(NaCl)=Λ0Na++Λ0Cl−Λ0m(NaCl)=50.1+76.3=126.4 S cm2 mol−1

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