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Standard XII

Chemistry

Kohlrausch Law

The values of...

Question

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:

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 B 271.44
$Λ_{e q}^{\circ} o r Λ_{(N H_{4} O H)}^{\infty} = Λ_{e q (N H_{4} C l)}^{\infty} + Λ_{e q (N a O H)}^{\infty} - Λ_{e q (N a C l)}^{\infty}$
$= (149.74 + 248.1 - 126.4)$
$= 271.44 o h m^{- 1} c m^{2} e q^{- 1}$

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Similar questions

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 :

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 value of

Ʌ_{e q}^{\infty}

for

N H_{4} C l

, NaOH and NaCl are respectively, 149.74, 248.1 and 126.4

Ω^{- 1} c m^{2} e q u i v^{- 1}

. The value of

Ʌ_{e q}^{\infty}

N H_{4} O H

Q. 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}^{-}

N H_{4} O H

is?

Q. Given:

Λ^{\infty} (\frac{1}{3} A l^{3 +}) = 63 Ω^{- 1} c m^{2} e q^{- 1}

and

Λ^{\infty} (\frac{1}{2} S O_{4}^{2 -}) = 80 Ω^{- 1} c m^{2} e q^{- 1}

. Then the value

Λ^{\infty} (A l_{2} (S O_{4})_{3})

would be:

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The values of Λ∞eq for NH4Cl,NaOH and NaCl are, respectively, 149.74, 248.1, and 126.4 ohm−1cm2eq−1. The value of Λ∞eqNH4OH is:

The correct option is B 271.44Λ∘eqorΛ∞(NH4OH)=Λ∞eq(NH4Cl)+Λ∞eq(NaOH)−Λ∞eq(NaCl)=(149.74+248.1−126.4)=271.44 ohm−1cm2eq−1

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:

The correct option is B 271.44
$Λ_{e q}^{\circ} o r Λ_{(N H_{4} O H)}^{\infty} = Λ_{e q (N H_{4} C l)}^{\infty} + Λ_{e q (N a O H)}^{\infty} - Λ_{e q (N a C l)}^{\infty}$
$= (149.74 + 248.1 - 126.4)$
$= 271.44 o h m^{- 1} c m^{2} e q^{- 1}$