Question

benzyl acetate is one of the active components of oil of jasmine. if 0.125 g of the compound is added to 25.0 g of chloroform the boiling point of the solution is 61.82 C what is the molar mass of benzyl acetate.

Answer 1

We must apply:

dTb = Kb*m

solve for "m" which is molality and Kb is given in tables

m = mol solute / kg solvent

m = 0.125 g of Benzyl ACetate

then

m = 25 g of chloroform; change to kg = 0.025 kg of chloroform

by definition

molality = mol solute / kg solvnet = 0.0008323345 /0.025 = 0.03329338 molal

then

Kb = 3.88

Tb normal = 61.2

dTb = Kb*m

Solve for m

m = dTb/Kb

dTb = Tb mix - Tb normal = 61.82-61.2 = 0.62°c

The real value of

then

m = dTb/Kb = 0.62/3.88 = 0.15979mol of solute / kg solvent

then

mol solute = 0.15979*kg sovlent = 0.15979*0.025 = 0.00399475 mol of solute

since

MW = mass/mol

and

MW = 0.125/0.00399475 = 31.291069 g/mol

NOTE that actual value is given as:

MW of B.Ac = 150.18 g/mol