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Standard XII

Chemistry

Conductance and Conductivity

The values of...

Question

The values of equivalent conductivity at infinite dilutions for $N H_{4} C l, N a O H$ and $N a C l$ are respectively $149.74, 248.1$ and $126.4$ $o h m^{- 1}$ $c m^{2}$ $e q u i^{- 1}$ . The value of $λ_{e q}^{-}$ of $N H_{4} O H$ is?

371.44

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271.44

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71.44

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It cannot be calculated from the data given

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Solution

The correct option is A $271.44$
According to Kohlrausch's law of equivalent conductivity, it could be stated:
$λ_{e q N H_{4} O H}^{-} = λ_{e q N H_{4} C l}^{-} + λ_{e q N a O H}^{-} - λ_{e q N a C l}^{-}$
$λ_{e q N H_{4} O H}^{-} = 149.74 + 248.1 - 126.4 = 271.44 o h m^{- 1} c m^{2} e q u i^{- 1}$

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Similar questions

Q. The conductivities at infinite dilution of

N H_{4} C l, N a O H

and

N a C l

are

130, 218, 120

o h m^{- 1}

c m^{2}

e q^{- 1}

. If equivalent conductance of

\frac{N}{100}

solution of

N H_{4} O H

10

o h m^{- 1}

c m^{2}

e q^{- 1}

, then degree of dissociation of

N H_{4} O H

at this dilution is:

The values $\land_{e q}^{\infty}$ for $N H_{4} C l, N a O H$ and $N a C l$ are 149.74, 248.1 and 126.4 $o h m^{- 1}$ $c m^{2}$ $e q v^{- 1}$ . The value of $\land_{e q}^{\infty}$ of $N H_{4} O H$ :

Q. The values of

Λ_{e q}^{\infty}

for

N H_{4} C l, N a O H

and

N a C l

are, respectively,

149.74

248.1,

and

126.4

o h m^{- 1} c m^{2} e q^{- 1} .

The value of

Λ_{e q}^{\infty} N H_{4} O H

is:

Q. The values of

Λ_{e q}^{0}

for

N H_{4} C l, N a O H

and

N a C l

are

149.74, 248.1

and

126.4 o h m^{- 1} c m^{2} e q^{- 1}

respectively. The value of

Λ_{e q}^{0} (N H_{4} O H)

is :

Q. At 291 K, the equivalent conductivities at infinite dilution of

N H_{4} C l

N a O H

and

N a C l

are

129.8

217.5

and

108.9

c m^{2} e q^{- 1}

respectively. The equivalent conductivity at infinite dilution of

N H_{4} O H

is:

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The values of equivalent conductivity at infinite dilutions for NH4Cl,NaOH and NaCl are respectively 149.74,248.1 and 126.4 ohm−1 cm2 equi−1. The value of λ−eq of NH4OH is?

The correct option is A 271.44According to Kohlrausch's law of equivalent conductivity, it could be stated:λ−eqNH4OH=λ−eqNH4Cl+λ−eqNaOH−λ−eqNaClλ−eqNH4OH=149.74+248.1−126.4=271.44ohm−1cm2equi−1

The values of equivalent conductivity at infinite dilutions for $N H_{4} C l, N a O H$ and $N a C l$ are respectively $149.74, 248.1$ and $126.4$ $o h m^{- 1}$ $c m^{2}$ $e q u i^{- 1}$ . The value of $λ_{e q}^{-}$ of $N H_{4} O H$ is?