Skip to content
dronacharya.ai
← JEE Advanced 2021 · Paper 2
ChemistryNumerical+2Question 27 of 57

JEE Advanced 2021 · Paper 2, Chemistry question 8

Passage

Question Stem for Question Nos. 7 and 8

At 298 K, the limiting molar conductivity of a weak monobasic acid is 4×1024 \times 10^2 S cm2^2 mol−1^{-1}. At 298 K, for an aqueous solution of the acid the degree of dissociation is α\alpha and the molar conductivity is y×102\mathbf y \times 10^2 S cm2^2 mol−1^{-1}. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y×1023\mathbf y \times 10^2 S cm2^2 mol−1^{-1}.

The value of y\mathbf y is ____.

Answer and worked solution

Official answer: 0.8

From the previous part, y×102=αΛ∘y\times10^2=\alpha\Lambda^\circ, so
y=400α100=4×1151≈0.86.y=\frac{400\alpha}{100}=4\times\frac{11}{51}\approx0.86.