The equivalent conductance of NaCl at concentration C and at infinite dilution are $\lambda_{\mathrm{C}}$ and $\lambda_{\infty}$, respectively. The correct relationship between $\lambda_{\mathrm{C}}$ and $\lambda_{\infty}$ is given as: (where the constant B is positive)

(A) $\lambda_{\mathrm{C}}=\lambda_{\infty}-(\mathrm{B}) \sqrt{\mathrm{C}}$

(B) $\lambda_{\mathrm{C}}=\lambda_{\infty}+(\mathrm{B}) \sqrt{\mathrm{C}}$

(C) $\lambda_{\mathrm{C}}=\lambda_{\infty}+(\mathrm{B}) \mathrm{C}$

(D) $\lambda_{\mathrm{C}}=\lambda_{\infty}-(\mathrm{B}) \mathrm{C}$

JEE Main · 2014 · 06 Apr 2014 · Offline · Redox Reactions and Electrochemistry

The equivalent conductance of NaCl at concentration C and at infinite dilution are

λ_{C}

and

λ_{\infty}

, respectively. The correct relationship between

λ_{C}

and

λ_{\infty}

is given as:
(where the constant B is positive)

(A) $λ_{C} = λ_{\infty} - (B) C$ ✓ Correct
(B) $λ_{C} = λ_{\infty} + (B) C$
(C) $λ_{C} = λ_{\infty} + (B) C$
(D) $λ_{C} = λ_{\infty} - (B) C$

Answer: (A)

Solution

For a strong electrolyte the Debye-Huckel-Onsager result says the molar conductance falls off with the square root of concentration as the ions begin to interfere with one another:

λ_{C} = λ_{\infty} - B C

, with

B

a positive constant. Conductance is highest at infinite dilution and drops as

C

rises.

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option

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JEE Main · 2014 · 06 Apr 2014 · Offline · Redox Reactions and Electrochemistry

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