Given the limiting molar conductivity as - 0m-HCl- - 425-9- - 1 cm2 mol - 1- 0m-NaCl- - 126-4- - 1 cm2 mol-1- 0m -CH3 COONa- - 91- - 1 cm2 mol- 1-The molar conductivity- at infinite dilution- of acetic -in - - - 1 cm2 mol-1- will be-390-5390- 299-5-481-5- 516-9-

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Byju's Answer

Standard XII

Chemistry

Kohlrausch Law

Given the lim...

Question

Given the limiting molar conductivity as
$^0 m (H C l) = 425.9 Ω - 1 c m 2 m o l - 1$
$^0 m (N a C l) = 126.4 Ω - 1 c m 2 m o l - 1$
$^0 m (C H 3 C O O N a) = 91 Ω - 1 c m 2 m o l - 1$
The molar conductivity, at infinite dilution, of acetic (in $Ω - 1 c m 2 m o l - 1$ ) will be:

481.5

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390.5390.

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299.5

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516.9

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Solution

The correct option is A 390.5
According to Kphlrush's law. Each ion makes a definite contribution to the total molar conductivity of an electrode irrespective of the nature of the other ion
eg for $A_{x} B_{y}$
$S_{m}^{o} = X_{λ (A^{y +})}^{o} + Y_{λ (B^{x -})}$
similarly
$Λ_{m (C H_{2} C O O H)}^{\infty} = Λ_{m (C H_{2} C O O H)}^{\infty} + Λ_{m (H C l)}^{\infty} - Λ_{m (N a C l)}^{\infty}$
$= 91 + 425.9 - 126.4 = 390 Ω^{- 1} c m^{2} m o l^{- 1}$

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Similar questions

Q. Given the limiting molar conductivity as

_{m}

(HCl)

= 425.9 Ω^{- 1} c m^{2} m o l^{- 1}

_{m}

(NaCl)

= 126.4 Ω^{- 1} c m^{2} m o l^{- 1}

_{m}

(CH3 COONa)

= 91 Ω^{- 1} c m^{2} m o l^{- 1}

The molar conductivity, at infinite dilution, of acetic acid (in

Ω^{- 1} c m^{2} m o l^{- 1}

) will be

Q. Molar conductance of

0.1 M

acetic acid is

7 {o h m}^{- 1} {c m}^{2} {m o l}^{- 1}

. If the molar conductance of acetic acid at infinite dilution is

380.8 {o h m}^{- 1} {c m}^{2} {m o l}^{- 1}

, the value of dissociation constant will be:

The molar conductances of NaCl, HCl and CH₃COONa at infinite dilution are 126.45, 426.16 and 91 ohm^-1 cm² mol^-1 respectively. The molar conductance of CH₃COOH at infinite dilution is

Q. Given the limiting molar conductivity as:

A_{m}^{\infty} (H C l) = 126.4 Ω^{- 1} {c m}^{2} {m o l}^{- 1}

A_{m}^{\infty} (N a C l) = 425.9 Ω^{- 1} {c m}^{2} {m o l}^{- 1}

A_{m}^{\infty} ({C H}_{3} C O O N a) = 91 Ω^{- 1} {c m}^{2} {m o l}^{- 1}

The molar conductivity at infinite dilution of acetic acid

(i n Ω^{- 1} {c m}^{2} {m o l}^{- 1})

will be:

Q. The molar ionic conductances at infinite dilution of

M g^{2 +}

and

C l^{-}

are 106.1 and 76.3

o h m^{- 1} o h m^{- 1} c m^{2} m o l^{- 1}

respectively. The Molar conductance of solution of

M g C l_{2}

at
infinite dilution is

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Given the limiting molar conductivity as ^0m(HCl)=425.9Ω−1cm2mol−1^0m(NaCl)=126.4Ω−1cm2mol−1^0m(CH3COONa)=91Ω−1cm2mol−1The molar conductivity, at infinite dilution, of acetic (in Ω−1cm2mol−1) will be:

Given the limiting molar conductivity as
$^0 m (H C l) = 425.9 Ω - 1 c m 2 m o l - 1$
$^0 m (N a C l) = 126.4 Ω - 1 c m 2 m o l - 1$
$^0 m (C H 3 C O O N a) = 91 Ω - 1 c m 2 m o l - 1$
The molar conductivity, at infinite dilution, of acetic (in $Ω - 1 c m 2 m o l - 1$ ) will be: