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Molar conductivities (Λm\Lambda^{\circ}_m) at infinite dilution of NaClNaCl, HClHCl, and CH3COONaCH_3COONa are 126.4126.4, 425.9425.9, and 91.0 S cm2 mol191.0\text{ S cm}^2\text{ mol}^{-1} respectively. Λm\Lambda^{\circ}_m for CH3COOHCH_3COOH will be:

A

180.5 S cm2 mol1180.5\text{ S cm}^2\text{ mol}^{-1}

B

290.8 S cm2 mol1290.8\text{ S cm}^2\text{ mol}^{-1}

C

390.5 S cm2 mol1390.5\text{ S cm}^2\text{ mol}^{-1}

D

425.5 S cm2 mol1425.5\text{ S cm}^2\text{ mol}^{-1}

Step-by-Step Solution

According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.

Λm(CH3COOH)=λH++λCH3COO\Lambda^{\circ}_m(CH_3COOH) = \lambda^{\circ}_{H^+} + \lambda^{\circ}_{CH_3COO^-}

This can be obtained by algebraically combining the given molar conductivities of strong electrolytes: Λm(CH3COOH)=Λm(CH3COONa)+Λm(HCl)Λm(NaCl)\Lambda^{\circ}_m(CH_3COOH) = \Lambda^{\circ}_m(CH_3COONa) + \Lambda^{\circ}_m(HCl) - \Lambda^{\circ}_m(NaCl) Λm(CH3COOH)=(91.0+425.9126.4) S cm2 mol1\Lambda^{\circ}_m(CH_3COOH) = (91.0 + 425.9 - 126.4)\text{ S cm}^2\text{ mol}^{-1} Λm(CH3COOH)=516.9126.4=390.5 S cm2 mol1\Lambda^{\circ}_m(CH_3COOH) = 516.9 - 126.4 = 390.5\text{ S cm}^2\text{ mol}^{-1}

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