see,
if u r writing the law in terms of molar conductivity then,
/\m0(BaCl2) = /\m0Ba+2 + 2 /\m0Cl -
if u r writing the law in terms of equivalent conductivity then,
/\eq.0(BaCl2) = /\eq.0Ba+2 + /\eq.0Cl -
Hope u got it . . .