Given the limiting molar conductivity as- -708-0m -160-HCl- - 425-9 -937-1 -160-cm2 -160-mol-1 -708-0m -160-NaCl- - 126-4 -937-1 -160-cm2 -160-mol-1 -708-0m -160-CH3 -160-COONa- - 91 -937-1 -160-cm2 -160-mol-1The molar conductivity at infinite dilution- of acetic acid -in -937-1 -160-cm2 -160-mol-1- will be- 516-5- 481-5- 390-5- 256-5-

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Equivalent, Molar Conductivity and Cell Constant

Given the lim...

Question

Given the limiting molar conductivity as:

˄0m (HCl) = 425.9Ω-1 cm2 mol-1
˄0m (NaCl) = 126.4Ω-1 cm2 mol-1
˄0m (CH3 COONa) = 91Ω-1 cm2 mol-1

The molar conductivity at infinite dilution, of acetic acid (in Ω-1 cm2 mol-1
) will be

481.5

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256.5

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390.5

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516.5

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Solution

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^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

Ω - 1 c m 2 m o l - 1

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}

(i n Ω^{- 1} {c m}^{2} {m o l}^{- 1})

_{m}

= 425.9 Ω^{- 1} c m^{2} m o l^{- 1}

_{m}

= 126.4 Ω^{- 1} c m^{2} m o l^{- 1}

_{m}

= 91 Ω^{- 1} c m^{2} m o l^{- 1}

Ω^{- 1} c m^{2} m o l^{- 1}

25^{o}

\land_{m}^{\infty}

426.2

{c m}^{2}

{m o l}^{- 1}

\land_{m}^{\infty}

{c m}^{2}

{m o l}^{- 1}

C H_{3} C O O K

\land_{m}^{\infty}

149.86

{c m}^{2}

{m o l}^{- 1}

25^{o}

145.0 S c m^{2} m o l^{- 1}

110.1 S c m^{2} m o l^{- 1}

A^{+}

73.5 S c m^{2} m o l^{- 1}

C^{+}

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