Given the limiting molar conductivity as -m HCl - 425-9-1 cm2 mol-1 -m NaCl - 126-4-1 cm2 mol-1 -m CH3 COONa - 91-1 cm2 mol-1The molar conductivity- at infinite dilution- of acetic acid in -1 cm2 mol-1 will be

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Standard X

Chemistry

Ionisation Energy

Given the lim...

Question

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

481.5

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390.5

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299.5

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516.9

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Solution

The correct option is B 390.5

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Similar questions

Q. 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:

Q. From the given molar conductivities infinite dilution, calculate

λ_{m}^{\infty}

for

N H_{4} O H

λ_{m}^{\infty} for B a (O H)_{2} = 457.6 o h m^{- 1} c m^{2} m o l^{- 1} .

λ_{m}^{\infty} for B a (C l_{2}) = 240.6 o h m^{- 1} c m^{2} m o l^{- 1} .

λ_{m}^{\infty} for N H_{4} C l = 128.8 o h m^{- 1} c m^{2} m o l^{- 1} .

Q. 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

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:

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Given the limiting molar conductivity as˚λm (HCl) =425.9Ω−1cm2mol−1˚λm (NaCl) =126.4Ω−1cm2mol−1˚λm (CH3 COONa) =91Ω−1cm2mol−1The molar conductivity, at infinite dilution, of acetic acid (in Ω−1cm2mol−1) will be

The correct option is B 390.5