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At 298 K, the molar conductivities at infinite dilution of sodium propionate (CH3CH2COONa), HCl and NaCl are 85.9, 426.1 and 126 ohm−1 cm2 mol−1 respectively. Calculate the molar conductivity of - Chemistry (Theory)

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Question

At 298 K, the molar conductivities at infinite dilution of sodium propionate (CH3CH2COONa), HCl and NaCl are 85.9, 426.1 and 126 ohm−1 cm2 mol−1 respectively. Calculate the molar conductivity of propionic acid at infinite dilution.

Numerical
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Solution

Given: `Lambda_m^0​` (CH3CH2COONa) = 85.9 Ω−1 cm2 mol−1

`Lambda_m^0` (HCl) = 426.1 Ω−1 cm2 mol−1

`Lambda_m^0` (NaCl) = 126.0 Ω−1 cm2 mol−1

To calculate the molar conductivity at infinite dilution `(Lambda_m^0)` of propionic acid (CH3CH2COOH), we use Kohlrausch’s Law of Independent Migration of Ions:

\[\ce{\Lambda{^{0}_{m}}​(CH3CH2COOH) = \Lambda{^{0}_{m}}(CH3CH2COONa) + \Lambda{^{0}_{m}}​(HCl) - \Lambda{^{0}_{m}}​(NaCl)}\]

= 85.9 + 426.1 − 126.0

= 386 Ω−1 cm2 mol−1

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Chapter 3: Electrochemistry - REVIEW EXERCISES [Page 171]

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Nootan Chemistry Part 1 and 2 [English] Class 12 ISC
Chapter 3 Electrochemistry
REVIEW EXERCISES | Q 3.56 | Page 171
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