AP EAMCET · Chemistry · Electrochemistry

The molar conductivities $(λ_{m}^{0})$ at infinite dilution of $KBr, HBr$ and ${KNH}_{2}$ are $120.5, 420.6$ and $90.48 S {cm}^{2} {mol}^{- 1}$ respectively. Find the value of $λ_{m}^{0}$ for ${NH}_{3}$ .

A $511 .0 S {cm}^{2} {mol}^{- 1}$
B $390.5 S {cm}^{2} {mol}^{- 1}$
C $256.2 S {cm}^{2} {mol}^{- 1}$
D $240.9 S {cm}^{2} {mol}^{- 1}$

Verified Solution

Answer & Solution

Correct Answer

(B) $390.5 S {cm}^{2} {mol}^{- 1}$

Step-by-step Solution

Detailed explanation

For KBr :- λKBr= λK+ + λBr-120.5=λK+ + λBr- For HBr :- λHBr= λH+ + λBr-420.6= λH+ + λBr- For KNH2 :-…

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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The molar conductivities λm0 at infinite dilution of KBr, HBr and KNH2 are 120.5,420.6 and 90.48 S cm2 mol-1 respectively. Find the value of λm0 for NH3.

Step-by-step Solution

The molar conductivities $(λ_{m}^{0})$ at infinite dilution of $KBr, HBr$ and ${KNH}_{2}$ are $120.5, 420.6$ and $90.48 S {cm}^{2} {mol}^{- 1}$ respectively. Find the value of $λ_{m}^{0}$ for ${NH}_{3}$ .