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Chemistry-(general)

NEET Chemistry (general) MCQ Question

Type: MCQ-numerical-Medium-Class 12

Calculate the dissociation constant KaK_a for acetic acid given that its conductivity is 4.95×10−5 S cm−14.95 \times 10^{-5} \text{ S cm}^{-1} at a concentration of 0.001028 mol L−10.001028 \text{ mol L}^{-1} and its limiting molar conductivity Λm0\Lambda_m^0 is 390.5 S cm2 mol−1390.5 \text{ S cm}^{2} \text{ mol}^{-1}.

A

1.78 × 10^-5 mol L^-1

B

0.1233 mol L^-1

C

48.15 S cm^2 mol^-1

D

4.95 × 10^-5 S cm^-1

Correct Answer

Option A

Detailed Explanation

Using the molar conductivity Λm=κ×1000c=48.15 S cm2 mol−1\Lambda_m = \frac{\kappa \times 1000}{c} = 48.15 \text{ S cm}^{2} \text{ mol}^{-1}, the degree of dissociation α\alpha is calculated as 0.12330.1233. The dissociation constant KaK_a is then 1.78×10−5 mol L−11.78 \times 10^{-5} \text{ mol L}^{-1} from Ka=cα21−αK_a = \frac{c\alpha^2}{1-\alpha}.

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