If the values of - of NH4Cl- NaOH and NaCl are 130- 217 and 109 ohm-1- cm2- equiv-1 respectively- the - of NH4OH in ohm-1- cm2- equiv-1 is-

The correct option is C 238 #160; ∧_∞ (NH_4OH) = ∧_∞ (NH_4Cl) + ∧_∞ (NaOH) ∧_∞ (NaCl) #160; #160; ∧_∞ (NH_4OH) = 130 + 217 109 #160; ...

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Byju's Answer

Standard XII

Physics

Infinite Line of Charge

If the values...

Question

If the values of $\land_{\infty}$ of $N H_{4} C l, N a O H$ and $N a C l$ are $130, 217$ and $109 o h m^{- 1} \cdot c m^{2} \cdot e q u i v^{- 1}$ respectively, the $\land_{\infty}$ of $N H_{4} O H$ in $o h m^{- 1} \cdot c m^{2} \cdot e q u i v^{- 1}$ is:

238

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198

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Solution

The correct option is C $238$
$\land_{\infty (N H_{4} O H)} = \land_{\infty (N H_{4} C l)} + \land_{\infty (N a O H)} - \land_{\infty (N a C l)}$

$\land_{\infty (N H_{4} O H)} = 130 + 217 - 109$

$\land_{\infty (N H_{4} O H)} = 238 o h m^{- 1} \cdot c m^{2} \cdot e q u i v^{- 1}$

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Similar questions

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 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. 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. 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:

Q. Equivalent conductance of

1

C H_{3} C O O H

10

o h m^{- 1}

c m^{2}

e q u i^{- 1}

and that at infinite dilution is

200

o h m^{- 1}

c m^{2}

e q u i v^{- 1}

. Hence the

%

ionization of

C H_{3} C O O H

is:

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If the values of ∧∞ of NH4Cl,NaOH and NaCl are 130,217 and 109 ohm−1⋅cm2⋅equiv−1 respectively, the ∧∞ of NH4OH in ohm−1⋅cm2⋅equiv−1 is:

The correct option is C 238 ∧∞(NH4OH)=∧∞(NH4Cl)+∧∞(NaOH)−∧∞(NaCl) ∧∞(NH4OH)=130+217−109 ∧∞(NH4OH)=238ohm−1⋅cm2⋅equiv−1

If the values of $\land_{\infty}$ of $N H_{4} C l, N a O H$ and $N a C l$ are $130, 217$ and $109 o h m^{- 1} \cdot c m^{2} \cdot e q u i v^{- 1}$ respectively, the $\land_{\infty}$ of $N H_{4} O H$ in $o h m^{- 1} \cdot c m^{2} \cdot e q u i v^{- 1}$ is:

The correct option is C $238$
$\land_{\infty (N H_{4} O H)} = \land_{\infty (N H_{4} C l)} + \land_{\infty (N a O H)} - \land_{\infty (N a C l)}$

$\land_{\infty (N H_{4} O H)} = 130 + 217 - 109$

$\land_{\infty (N H_{4} O H)} = 238 o h m^{- 1} \cdot c m^{2} \cdot e q u i v^{- 1}$