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Standard XII

Chemistry

Kohlrausch Law

If the molar ...

Question

If the molar conductivities (in S cm $^{2}$ mol $^{- 1}$ ) of NaCl, KCl and NaOH at infinite dilution are 126, 150 and 250 respectively, the molar conductivity of KOH is:
[in S cm $^{2}$ mol $^{- 1}$ ]

526

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226

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26

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274

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Solution

The correct option is D $274$
$λ_{K O H}$ $= λ_{K C l}$ $+ λ_{N a O H}$ $- λ_{N a C l}$

$λ_{K O H} = 150 + 250 - 126 = 274 S c m^{2} m o l^{- 1}$

Hence, the correct option is $D$

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If the molar conductivities (in S cm2 mol−1) of NaCl, KCl and NaOH at infinite dilution are 126, 150 and 250 respectively, the molar conductivity of KOH is:[in S cm2 mol−1]

The correct option is D 274λKOH =λKCl +λNaOH −λNaClλKOH=150+250−126=274 S cm2 mol−1Hence, the correct option is D

If the molar conductivities (in S cm $^{2}$ mol $^{- 1}$ ) of NaCl, KCl and NaOH at infinite dilution are 126, 150 and 250 respectively, the molar conductivity of KOH is:
[in S cm $^{2}$ mol $^{- 1}$ ]

The correct option is D $274$
$λ_{K O H}$ $= λ_{K C l}$ $+ λ_{N a O H}$ $- λ_{N a C l}$

$λ_{K O H} = 150 + 250 - 126 = 274 S c m^{2} m o l^{- 1}$

Hence, the correct option is $D$