The values of the limiting molar conductivity - for NaCl- HCl and NaOAc are 126-4- 425-9 and 91-0 S cm2 mol-1- respectively- For HOAc- - in S cm2 mol-1 is-

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Standard XII

Chemistry

Equivalent, Molar Conductivity and Cell Constant

The values of...

Question

The values of the limiting molar conductivity $(\land^{\circ})$ for $N a C l, H C l$ and $N a O A c$ are $126.4, 425.9$ and $91.0 S c m^{2} m o l^{- 1}$ , respectively. For $H O A c, \land^{\circ}$ in $S c m^{2} m o l^{- 1}$ is:

390.5

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299.5

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208.5

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217.4

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Solution

The correct option is D $390.5$
From the given data we can write the equation,

$λ_{H O A c} = λ_{H C l} + λ_{N a O A c} - λ_{N a C l}$

$λ_{H O A c} = 425.9 + 91 - 126.4 = 390.5 S c m^{2} m o l^{- 1}$

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Similar questions

Q. Molar conductivities at infinite dilution of

N a C l

H C l

and

C H_{3} C O O N a

are 126.4, 425.9 and 91.0

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

respectively. Molar conductivity for

C H_{3} C O O H

will be:

Q. The limiting molar conductivities of

N a C l

K B r

and

K C l

are 126, 152 and 150

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

respectively. The limiting molar conductivity for

N a B r

is:

Electrolyte	$K C l$	$K {N O}_{3}$	$H C l$	$N a O A c$	$N a C l$
$Λ^{\infty} (S {c m}^{2} {m o l}^{- 1})$	$149.9$	$145.0$	$426.2$	$91.0$	$126.5$

Calculate

{Λ^{\infty}}_{H O A c}

using appropriate molar conductances of the electrolytes listed above at infinite dilution in

H_{2} O

25^{o} C

Electrolyte $\land^{\infty} (S c m^{2} m o l^{- 1})$

$K C l$ 149.9

$K N O_{3}$ 145.0

$H C l$ 426.2

$N a O A c$ 91.0

$N a C l$ 126.5

Calculate $Λ {^{\infty}}_{H O A c}$ using appropriate molar conductances of the electrolytes listed above at infinite dilution with $H_{2} O$ at 25 $^{o}$ C?

Q. The limiting molar conductivities

Λ^{0}

for

N a C l, K B r and K C l

are

126, 152 and 150 S c m^{2} m o l^{- 1}

respectively. The

Λ^{0}

for

N a B r

is:

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The values of the limiting molar conductivity (∧∘) for NaCl,HCl and NaOAc are 126.4,425.9 and 91.0 S cm2mol−1, respectively. For HOAc,∧∘ in S cm2mol−1 is:

The correct option is D 390.5From the given data we can write the equation,λHOAc=λHCl+λNaOAc−λNaClλHOAc=425.9+91−126.4=390.5 S cm2 mol−1

The values of the limiting molar conductivity $(\land^{\circ})$ for $N a C l, H C l$ and $N a O A c$ are $126.4, 425.9$ and $91.0 S c m^{2} m o l^{- 1}$ , respectively. For $H O A c, \land^{\circ}$ in $S c m^{2} m o l^{- 1}$ is:

The correct option is D $390.5$
From the given data we can write the equation,

$λ_{H O A c} = λ_{H C l} + λ_{N a O A c} - λ_{N a C l}$

$λ_{H O A c} = 425.9 + 91 - 126.4 = 390.5 S c m^{2} m o l^{- 1}$