Molar conductances of `BaCl_(2), H_(2)SO_(4)` and `HCl` at infinite dilutions are `X_(1), x_(2)` and `X_(3)` respectively. Equivalent conductance of `BaSO_(4)` at infinite dilution is :

Options

A. `([x_(1)+x_(2)-x_(3)])/(2)`
B. `[(x_(1)-x_(2)-x_(3)])/(2)`
C. `2(x_(1)+x_(2)-2x_(3))`
D. `([x_(1)+x_(2)-2x_(3)])/(2)`

Answer

D

Explanation

`^^_(m_(i)BaSO_(4)=(x_(1)+x_(2)x-2x_(3))implies^^_(eq.BaSO_(4))=(^^_(eq.BaSO_(4))/(n-"factor")`
`^^_(eq.BaSO_(4))=((x_(1)+x_(2)-2x_(3)))/(2)`