A 5 . 0 m mol dm - 3 aqueous solution of KCl has a conductance of 0 . 55 mS when measured in a cell constant 1 . 3 cm - 1 . The molar conductivity of this solution is _______ mSm 2 mol - 1 . (Round off to the Nearest Integer)

Step-by-step solution

Given conc n of KCl = m . mol L : Conductance G = 0 . 55   mS : Cell constant ℓ A = 1 . 3   cm - 1 To Calculate : Molar conductivity λ m of sol. → Since Λ m = 1 1000 × k M . . . ( 1 ) → Molarity = 5 × 10 - 3 mol L → Conductivity = G × ℓ A = 0 . 55   mS × 1 . 3 1 100   m - 1 = 55 × 1 . 3   mSm - 1 eq n   1 Λ m = 1 1000 × 55 × 1 . 3 5 1000 mSm 2   mol ⇒ Λ m = 14 . 3 mSm 2   mol