Question

Assume you dissolve 41.4 g of camphor (C10H16O) in 449 mL of ethanol, C2H5OH. Calculate the molality, mole fraction, and weight percent of camphor in this solution. (The density of ethanol is 0.785 g/mL.)

Answer 1

mass of camphor = 41.4 g

mass of ethanol = 449 x 0.785 = 352.5 g

1) molality = (W /MW) (1000 / mass of solvent in g)

W = 41.4 g

MW = 152.23 g/mol

molality = (41.4 / 152.23) (1000 / 352.5)

molality = 0.771 m

2) mole fraction = moles of camphor / total moles

moles of camphor = 41.4 / 152.23 = 0.272

moles of ethanol = 352.5 / 46.07 = 7.64

total moles = 0.272 + 7.64 = 7.912

mole fraction = 0.272 / 7.912

mole fraction = 0.0343

3) % mass = (mass of camphor / total mass) x 100

% mass = (41.4 / 41.4 + 352.5) x 100

% mass = (41.4 / 393.9) x 100

% mass = 10.51 %