Question

(6 pt) 4 A 1.07-mg sample of a non-ionic compound was dissolved in 78.1 mg of camphor, a solvent that melts at 179.5°C and has an unusually high Kr of 40.0 mola If the freezing point of the solution is 176.0°C: a) Calculate AT, then determine the molality (m) of this solution: b) Calculate the moles of solute in this solution: e) Calculate the molar mass ()of e coound:

Answer 1

a)

∆T = Freezing point of solvent - Freezing point of solution

∆T = 179.5℃ - 176℃ = 3.5℃

∆T= K_f × b

b = molality of solute

K_f= freezing point depression constant , 40℃/m

3.5℃ = 40℃/m × b

b = 3.5℃/ 40℃/m

b = 0.0875m

b)

molality = number of moles per kilogram of solvent

0.0875m = 0.0875moles of solute per kilogram of solvent

mass of camphor = 78.1mg = 0.0781g

number of moles of the compound for 0.0781g of camphor = (0.0875mol/1000g)×0.0781g = 6.834×10^-6mol

c)

molar mass = mass/no of mol

Number of moles of the compound = 6.834×10^-6mol

mass of the compound = 1.07mg = 0.00107g

molar mass of the compound = 0.00107g/6.834×10^-6mol = 156.6g/mol