Question

You want to prepare an aqueous solution of glycerol, C₃H₅(OH)₃, in which the mole fraction of the solute is .093. What mass of glycerol must you add to 425g of water to make this solution? What is the molality of the solution?

Answer 1

1)

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 425 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(4.25*10^2 g)/(18.02 g/mol)

= 23.59 mol

Mole fraction of water = 1 - mole fraction of glycerol

= 1 - 0.093

= 0.907

Mole fraction of water = mol of H2O / (mol of H2O + mol of glycerol)

0.907 = 23.59 / ( 23.59 + mol of glycerol)

23.59 + mol of glycerol = 26.01

mol of glycerol = 2.42 mol

Molar mass of C3H5(OH)3,

MM = 3*MM(C) + 8*MM(H) + 3*MM(O)

= 3*12.01 + 8*1.008 + 3*16.0

= 92.094 g/mol

use:

mass of C3H5(OH)3,

m = number of mol * molar mass

= 2.42 mol * 92.09 g/mol

= 2.229*10^2 g

Answer: 223 g

2)

m(solvent)= 425 g

= 0.425 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(2.42 mol)/(0.425 Kg)

= 5.694 molal

Answer: 5.69 molal