Molar conductance 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 will be:-

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

`lambda_(m(BaSO_(4)))^(oo) = lambda_(Ba^(2+))^(oo) + lambda_(SO_(4)^(2-))^(oo)`
`lambda_(m(BaCl_(2)))^(oo) + lambda_(m(h_(2)SO_(4)))^(oo) - 2lambda_(m(HCl))^(oo)`
`= x_(1) + x_(2) - 2x_(3)`
and `lambda_(eq.)^(oo) = (1)/(2)lambda_(m(BaSO_(4)))^(oo) = (1)/(2)(x_(1) + x_(2) + x_(3))`