Molar conductivities of `BaCl_(2), H_(2)SO_(4) and HCl` at infinite dilution are `x_(1) and x_(2), and x_(3)` respectively. Molar conductivity of `BaSO_(4)` at infinite dilution will be

Options

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

Answer

B

Explanation

Using Kohlrausch.s law, `Lamda_(mBaCl_(2)) = lamda^(@) Ba^(2+) + 2lamda^(@) Cl^(-) = x_(1)` ..(1)
`Lamda_(mH_(2)SO_(4))^(@)= 2lamda^(@)H^(+) + lamda^(@) SO_(4)^(2-) = x_(2)` …(2)
`Lamda_(mHCl)^(@)= lamda^(@) H^(+) + lamda^(@) Cl^(-) = x_(3)` ...(3)
Adding the Eqs (1) and (2) and subtracting `2 xx` Eq (3), we get `Lamda_(mBeSO_(4))^(@)= x_(1) + x_(2)-2x_(3)`