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0.90 g of an  1:2 electrolyte was dissolved in 87.90 g of benzene. This raised the boiling point of benzene by 0.250 C. If the molecular mass of the electrolyte is 103.0molar calculate the molal elevation constant for benzene. Given that the solute dissociate with degree of dissociation 0.25.
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM

ΔTb = i Kbm

α =   i -1 / n-1

0.25 = i-1/ 3-1     i-1 = 0.50   i = 1.50
0.25 = [1.50 x Kb x 0.90/ 103] x 1000 / 87.90
= 1.50 x Kb x 0.0087 x 1000 /87.90
= 1310.67 Kb  / 87.90 = 14.90
Kb = 0.25/14.90= 0.0167 K Kg mol-1

Answered by | 04 Jun, 2014, 03:23: PM

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