The limiting molar conductivities of the given electrolytes at 298 K follow the order [\lambda^0 (K+ ) = 73.5, \lambda^ 0 (Cl^{–} ) = 76.3, \lambda^ 0 (Ca^{2+}) =119.0, \lambda^ 0 (SO ) ^{2–} _{4} = 160.0 \S \cm^2 \mol^{–1}{]}

Correct Answer is : KCl<CaCl2<K2SO4

KVPY SX SB Exam 19-Nov-2017 Question Paper

Question : 60 of 120

Marks: +1, -0

The limiting molar conductivities of the given electrolytes at 298 K follow the order

[λ^0 (K+ ) = 73.5, λ^ 0 (Cl^{–} ) = 76.3,

λ^ 0 (Ca^{2+}) =119.0, λ^ 0 (SO ) ^{2–} _{4} = 160.0 \S \cm^2 \mol^{–1}{]}

Solution:

λ°_\KCl= 73.5 + 76.3 = 149.8

λ°_{\CaCl _2}= 119 +2 × 76.3 = 271.6

λ°_{\K_2SO_4}=2 × 73.5 + 160 = 307

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