1.

Molar conductivity of KCl, NaCl and KNO_(3) are 150, 126 and 109 SC m^(2)mol^(-1) respectively, then what is the molar conductivity of NaNO_(3) ?

Answer»

385 S `cm^(2)mol^(-1)`
133 S `cm^(2)mol^(-1)`
167 S `cm^(2)mol^(-1)`
85 S `cm^(2)mol^(-1)`

Solution :`Lamda_(NaNO_(3))^(@)=Lamda_(NaCl)^(@)+Lamda_(KNO_(3))^(@)-Lamda_(KCL)^(@)`
`=126-109-150`
=85 S `cm^(2)mol^(-1)`


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