a solution prepared by dissolving 1.25g of oil of winter green in 99gof benzene has its bp of 80.31o C .determne the molar massofthis component.bp of benzene=80.1oC ,kb =2.53 Share with your friends Share 21 Avni answered this ∆Tb = 80.31 - 80.10 = 0.21oC = 0.21 KGiven mass of solute (WB) = 1.25 gKb = 2.53oC kg mol-1Molar mass of solute (MB) = ?Given mass of solvent (WA) = 99 gMB = Kb×WB×1000∆Tb×WAMB= 2.53×1.25×10000.21×99 = 152 gmol-1 49 View Full Answer