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← JEE Advanced 2026 · Paper 2
ChemistrySingle correct+3 / -1Question 37 of 54

JEE Advanced 2026 · Paper 2, Chemistry question 1

At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below:

Salt Concentration (M) Molar conductivity (S cm2^2 mol−1^{-1})
NaNOX3\ce{NaNO3} 0.01 111
0.04 101
NaCl\ce{NaCl} 0.01 117
0.04 107
AgNOX3\ce{AgNO3} 0.01 125
0.04 116

The conductivity of a saturated aqueous solution of AgCl is 1.40×10−61.40\times10^{-6} S cm−1^{-1} at 300 K. If the solubility of AgCl in water at 300 K is X mol L−1^{-1}, then log⁡10(X−1)\log_{10}(X^{-1}) is

(Assume that AgCl dissolved in water ionizes completely and that the molar conductivity of saturated AgCl solution is equal to its limiting molar conductivity.)

Choose one option
Answer and worked solution

Official answer: C

Limiting molar conductivities. For strong electrolytes Λ=Λ∘−Ac\Lambda=\Lambda^\circ-A\sqrt c. With 0.01=0.1\sqrt{0.01}=0.1 and 0.04=0.2\sqrt{0.04}=0.2, each salt loses 10 or 9 units per 0.1 in c\sqrt c:

  • Λ∘(NaNOX3)=111+10=121\Lambda^\circ(\ce{NaNO3})=111+10=121
  • Λ∘(NaCl)=117+10=127\Lambda^\circ(\ce{NaCl})=117+10=127
  • Λ∘(AgNOX3)=125+9=134\Lambda^\circ(\ce{AgNO3})=125+9=134

(all in S cm² mol⁻¹).

Kohlrausch.
Λ∘(AgCl)=134+127−121=140 S cm2 mol−1.\Lambda^\circ(\ce{AgCl})=134+127-121=140\ \text{S cm}^2\text{ mol}^{-1}.

Solubility.
X=1000 κΛ∘=1000×1.40×10−6140=10−5 mol L−1,X=\frac{1000\,\kappa}{\Lambda^\circ}=\frac{1000\times1.40\times10^{-6}}{140}=10^{-5}\ \text{mol L}^{-1},
so log⁡10(X−1)=5\log_{10}(X^{-1})=5.

Topics:Conductance and Kohlrausch's lawSolubility product
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