An aqueous solution containing 12 48g of barium chloride in 1 0kg of water boils at 373 0832K Calculate - Chemistry - Solutions

Question

An aqueous solution containing 12 48g of barium chloride in 1 0kg
of water boils at 373 0832K Calculate the degree of dissociation of
barium chloride (Given Kb for water= 0 52K/m)

Geetha · Accepted Answer

According to elevation of boiling point,∆Tb = Kb × m     = Kb × W2M2 × W1 ∆Tb - elevation of boiling pointKb - molal elevation constantW2 - Weight of the soluteM2 - Molecular mass of the soluteW1 - Weight of the solvent∆Tb  = Kb × W2M2 × W1 × 1000    = 0
52 × 12 48208
23× 1     = 0
0312∆Tb (given) = Tb - Tb0 = 373
0832 - 373 = 0
0832Degree of dissociationα = i-1n -1Where i = Vant Hoff factori = Observed colligative propertyCalculated colligative property  = 0
08320 0312 = 2 67α = 2
67-13 -1   (where n = 3 as BaCl2 dissociates BaCl2 → Ba2+ + 2Cl-   = 1
672 = 0 8333   = 0
8333 × 100 = 83
33 %

An aqueous solution containing 12.48g of barium chloride in 1.0kg of water boils at 373.0832K. Calculate the degree of dissociation of barium chloride. (Given Kb for water= 0.52K/m)