From the following molar conductivities at infinite dilution calculate limiting molar conductivity for NH4OH.... Limiting molar conductivity for Ba(OH)2=457.6 ohm-1 cm2 mol-1 Limiting molar conductivity for Bacl2=240.6 ohm-1cm-1 Limiting molar conductivity for NH4Cl=129.8 ohm-1cm2 mol-1 Share with your friends Share 19 Geetha answered this The formation of BaCl2 is represented asBa(OH)2 +2 NH4Cl →2NH4OH + BaCl2λMBa(OH)2 ∞ +2 λMNH4Cl ∞ = 2 λMNH4OH ∞ + λMBaCl2 ∞ 457.6 + 2 × 129.8 = 2 × λMNH4OH ∞ + 240.62 λMNH4OH ∞ =476.6λMNH4OH ∞ =476.62 =238.3 ohm-1cm2mol-1 22 View Full Answer Kanishka Yadav answered this limiting molar conductivity for NH4OH =LMC for (NH4^+) +(OH^-) ADDING bacl2 in the solution now limiting molar conductivity [ NH4CL+1/2 Ba(OH)2 -Bacl2] substitute the values u will get ans as 118 ohm-1 cm2 mol-1 -1
