Question

1.02g er urea (molare mass=60 g/mol) when dissolved in 98.5gram of certain solvent deceases the Freizing point by 0.211* 1,609 gram unknown compound when dissolved in 86 gram seme solvent depresses the frizing Point by 0.34* caculate the molare mass of unknown compound​

Answer 1

70 g          

Explanation:

Given: mass of urea = 1.02 g, ΔTf = 0.211. mass of unknown M = ?

number of moles of urea = 1.02/60 = 0.017

Molality = number of moles × 1000/98.5

              = 0.017 × 10.15

              = 0.1725

ΔTf₁ = 0.211 = Kf × 0.1725

ΔTf = Kd × 1.60/M × 1000/80.6 = 17.59/M

ΔTf₂ = 0.34 = Kf × 19.85/M

M = (0.211/0.34) × (19.85/17.59)

   = 70 g