The molar conductivities of KCl, NaCl and `KNO_(3)` are 152,128 and `111" S cm"^(2) mol^(-1)` respectively. What is the molar conductivity of `NaNO_(3)` ?

Options

A. 101 S `cm^(2) mol^(-1)`
B. `87 S cm^(2) mol^(-1)`
C. `-101 S cm^(2) mol^(-1)`
D. `-391 S c^(2) mol^(-1)`

Answer

B

Explanation

`Lambda_(NaNO_(3))^(@)=Lambda_(NaCl)^(@)+Lambda_(KNO_(3))^(@)-Lambda_(KCl)^(@)`
`=128+111-152=87 S cm^(2) mol^(-1)`