JEE Advanced · Chemistry · 17. Electrochemistry

The conductance of $0 .0015 M$ aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized $Pt$ electrodes. The distance between the electrodes is $120 cm$ with an area of cross-section of $1 {cm}^{2}$ . The conductance of this solution was found to be $5 \times 10^{- 7} S$ . The $pH$ of the solution is $4$ . The value of limiting molar conductivity of this weak monobasic acid in aqueous solution is $Z \times 10^{2} S {cm}^{- 1} {mol}^{- 1} .$ The value of $Z$ is.

A 3
B 6
C 9
D 12

Verified Solution

Answer & Solution

Correct Answer

(B) 6

Step-by-step Solution

Detailed explanation

For weak acid,

[H^{+}] = cα = c \times \frac{\land_{m}^{c}}{\land_{m}^{\infty}}

= c \times \frac{k \times \frac{1000}{c}}{\land_{m}^{\infty}} = \frac{k \times 1000}{\land_{m}^{\infty}} = \frac{G \times (\frac{l}{a}) \times 1000}{\land_{m}^{\infty}}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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(B)	(Q)
(C)	(R)
(D)	(S)
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