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Question

If the molar conductance values of Ca2+ and Cl at infinite dilution are respectively 118.88×104m2 mho mol1 and 77.33×104m2mho mho1 then that of CaCl2 is:

A
118.88×104
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B
154.66×104
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C
273.54×104
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D
196.21×104
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Solution

The correct option is C 273.54×104
mCaCl2=λCa2++2λCl
=(118.88××104)+2(77.33×104
=273.54×104m2mho mol1.

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