Chemistry, asked by tabbharmal8086, 9 months ago

Limiting molar conductivities for NaCl, HCl and CH3COONarespectively are 126.4, 425.9 and 91.0 S cm2 mol-1. Limitingmolar conductivity for CH3COOH is690.5 S cm2 mol-1590.5 S cmmol-1390.5 S cm2 mol-1490.5 S cm2 mol-1

Answers

Answered by priyadave2007

3

Answer:

CH3COONa+HCl⟶NaCl+CH3COOH

⇒91+425.9=126.4+x

⇒x=516.9−126.4

=390.5Scm2mol−1

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