Calculate - of CH3COOH if -m for HCl- NaCl- CH3COONa are 426- 126- 91 S Cm1- respectively- and -m - 14-4 S cm2 mol-1 at 0-015 M concentration-

The correct option is A 0.037 As given, Λ^∘_m(CH_3COOH) nbsp;= (426) 126 + 91 = 391 S cm^2 mol^ 1 = α = nbsp;14.4/391 ≈ 0 ...

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Byju's Answer

Standard XII

Chemistry

Nernst Equation

Calculate α...

Question

Calculate $α$ of $C H_{3} C O O H$ if $Λ_{m}^{\infty}$ for $H C l, N a C l$ , $C H_{3} C O O N a$ are $426, 126, 91$ S $C m^{1}$ , respectively, and $Λ_{m} = 14.4 S c m^{2} m o l^{- 1}$ at $0.015$ M concentration.

0.037

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0.018

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Solution

The correct option is A $0.037$
As given,
$Λ_{m (C H_{3} C O O H)}^{\circ} = (426) - 126 + 91$
$= 391 S c m^{2} m o l^{- 1}$
$= α = \frac{14.4}{391} \approx 0.037$

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Calculate α of CH3COOH if Λ∞m for HCl,NaCl, CH3COONa are 426,126,91 S Cm1, respectively, and Λm=14.4Scm2mol−1 at 0.015 M concentration.

The correct option is A 0.037As given,Λ∘m(CH3COOH)=(426)−126+91=391Scm2mol−1=α=14.4391≈0.037

Calculate $α$ of $C H_{3} C O O H$ if $Λ_{m}^{\infty}$ for $H C l, N a C l$ , $C H_{3} C O O N a$ are $426, 126, 91$ S $C m^{1}$ , respectively, and $Λ_{m} = 14.4 S c m^{2} m o l^{- 1}$ at $0.015$ M concentration.

The correct option is A $0.037$
As given,
$Λ_{m (C H_{3} C O O H)}^{\circ} = (426) - 126 + 91$
$= 391 S c m^{2} m o l^{- 1}$
$= α = \frac{14.4}{391} \approx 0.037$