Limiting molar conductivity of {\NH}_4 {\OH}(. i.e. \\Lambda^{0}_{ {\m}}( {\NH}_4 {\OH})) is equal to:

Correct Answer is : 0Λm(NH4Cl)+0Λm(NaOH)−0Λm(NaCl)

\stackrel{0}{\\Lambda}_{ {\m}}( {\NH}_4 {\Cl})+\stackrel{0}{\\Lambda}_{ {\m}}( {\NaCl})-\stackrel{0}{\\Lambda}_{ {\m}}( {\NaOH})

\stackrel{0}{\\Lambda}_{ {\m}}( {\NaOH})+\stackrel{0}{\\Lambda}_{ {\m}}( {\NaCl})-\stackrel{0}{\\Lambda}_{ {\m}}( {\NH}_4 {\Cl})

\stackrel{0}{\\Lambda}_{ {\m}}( {\NH}_4 {\OH})+\stackrel{0}{\\Lambda}_{ {\m}}( {\NH}_4 {\Cl}) -\stackrel{0}{\\Lambda}_{\m}( {\HCl})

\stackrel{0}{\\Lambda}_{ {\m}}( {\NH}_4 {\Cl})+\stackrel{0}{\\Lambda}_{ {\m}}( {\NaOH})-\stackrel{0}{\\Lambda}_{ {\m}}( {\NaCl})

AIPMT 2012 Physics and Chemistry Paper

Question : 97 of 100

Marks: +1, -0

Limiting molar conductivity of NH4‌OH(. i.e.

m(NH4‌OH)) is equal to:

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