KVPY SX SB Exam 20141 Question Paper

Question : 53 of 120

Marks: +1, -0

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 (in S cm

^{2 }

mol

^{-1}

) is

Solution:

126 =λ^∞_{\Na^{+}}+λ^∞_{\Cl^{-}}

150 =λ^∞_{\K^{+}}+λ^∞_{\Cl^{-}}

250 =λ^∞_{\Na^{+}}+λ^∞_{\OH^{-}}

_________________________________

(150+ 250 -126)=λ^∞_{\K^{+}}+λ^∞_{\OH^{-}}

λ^∞_{\KOH }=274

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