- of NH4OH at infinite dilution is S cm2 mol-1 given - of OH-174- if Cl-66 and - of NH4Cl-130S cm2mol-1

The correct option is C 238 λ #160; ^∞ #160; _ NH _ 4 OH =λ #160; ^∞ #160; _ NH _ 4 ^ + +λ #160; ^∞ #160; _ OH ^( ) #160; =( λ ...

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Byju's Answer

Standard XII

Chemistry

Kohlrausch Law

λ∞ of NH4OH...

Question

$λ^{\infty}$ of ${N H}_{4} O H$ at infinite dilution is ______ $S$ ${c m}^{2}$ ${m o l}^{- 1}$ given $λ^{\infty}$ of ${O H}^{-} = 174, λ^{\infty}$ if ${C l}^{-} = 66$ and $λ^{\infty}$ of ${N H}_{4} C l = 130 S$ ${c m}^{2} {m o l}^{- 1}$

140

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238

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Solution

The correct option is C $238$
${λ^{\infty}}_{{N H}_{4} O H} = {λ^{\infty}}_{{N H}_{4}^{+}} + {λ^{\infty}}_{{O H}^{(-)}}$
$= ({λ^{\infty}}_{{N H}_{4} C l} - {λ^{\infty}}_{{C l}^{(-)}}) + {λ^{\infty}}_{{O H}^{(-)}}$
$= 130 - 66 + 174$
$= 238$ ${s c m}^{2} {m o l}^{- 1}$

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Similar questions

Q. The degree of dissociation of 2.5 × 10^–2 M methanoic acid solution having molar conductivity 46.1 S cm² mol^–1 will be (Given : λ°(H⁺) = 349.6 S cm² mol^–1 and λ°(HCOO^–) = 54.6 S cm² mol^–1)

मोलर चालकता 46.1 S cm² mol^–1 वाले 2.5 × 10^–2 M मेथनॉइक अम्ल विलयन के वियोजन की मात्रा होगी (दिया है: λ°(H⁺) = 349.6 S cm² mol^–1 तथा λ°(HCOO^–) = 54.6 S cm² mol^–1)

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and % dissociation of 0.01 M

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Q. (a) The conductivity of 0.001

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and

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426.2

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KCl:

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= 114.42 S

{c m}^{2}

{m o l}^{- 1}

C H_{3} C O O K

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149.86

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λ∞ of NH4OH at infinite dilution is ______ S cm2 mol−1 given λ∞ of OH−=174,λ∞ if Cl−=66 and λ∞ of NH4Cl=130S cm2mol−1

The correct option is C 238λ∞NH4OH=λ∞NH+4+λ∞OH(−)=(λ∞NH4Cl−λ∞Cl(−))+λ∞OH(−)=130−66+174=238 scm2mol−1

$λ^{\infty}$ of ${N H}_{4} O H$ at infinite dilution is ______ $S$ ${c m}^{2}$ ${m o l}^{- 1}$ given $λ^{\infty}$ of ${O H}^{-} = 174, λ^{\infty}$ if ${C l}^{-} = 66$ and $λ^{\infty}$ of ${N H}_{4} C l = 130 S$ ${c m}^{2} {m o l}^{- 1}$

The correct option is C $238$
${λ^{\infty}}_{{N H}_{4} O H} = {λ^{\infty}}_{{N H}_{4}^{+}} + {λ^{\infty}}_{{O H}^{(-)}}$
$= ({λ^{\infty}}_{{N H}_{4} C l} - {λ^{\infty}}_{{C l}^{(-)}}) + {λ^{\infty}}_{{O H}^{(-)}}$
$= 130 - 66 + 174$
$= 238$ ${s c m}^{2} {m o l}^{- 1}$