Question

Mole fraction and mass percent:

1) An aqueous solution is36.0 % by masspotassium bromide,KBr, and has a density of 1.33 g/mL.

The mole fraction ofpotassium bromide in the solution is .

2) The mole fraction ofammonium carbonate,(NH₄)₂CO₃, in an aqueous solution is6.61×10^-2 .

The percent by mass ofammonium carbonate in the solution is  %.

Answer 1

1) An aqueous solution is36.0 % by masspotassium bromide,KBr, and has a density of 1.33 g/mL.

The mole fraction ofpotassium bromide in the solution is .

Because it says 36.0% BY MASS, it means that KBr is 36.0% of the total mass of the solution.

Density = mass / volume

Mass = density * volume

volume = 1 L or 1000 ml

density = 1.33 g/ml

mass = 1.33 g/ml x 1000 ml
mass = 1330 g

So now mass of KBr in 1 L
= 36.0% of 1330 g
= 0.36 x 1330 g
= 478.8 g of KBr

Now moles KBr = mass / molar mass
moles KBr = 478.8 g / 119.00 g/mol
= 4.02 moles

Thus mass water = 1330 g – 478.8g
= 851.2 g

Now moles H2O = mass / molar mass
moles H2O = 851.2 g / 18.02 g/mol
= 47.24moles

The mole fraction ofpotassium bromide = Moles of KBr/ total moles

= 4.02 /4.02+47.24

=4.02/51.26

=0.078

2) The mole fraction ofammonium carbonate,(NH₄)₂CO₃, in an aqueous solution is6.61×10^-2 .

The percent by mass ofammonium carbonate in the solution is  %.

Given that;

The mole fraction ofammonium carbonate,(NH₄)₂CO₃, in an aqueous solution is6.61×10^-2

6.61 x 10^-2 moles ,(NH₄)₂CO₃/ 1.00

1.00 = 6.61 x 10^-2 + x
x = moles water = 0.9339

mass ,NH₄)₂CO₃= 6.61 x 10^-2 mol x 96.09 g/mol= 6.34 g
Mass water = 0.9339 mol x 18.02 g/mol= 16.8 g

mass solution = 16.8 + 6.34= 23.14g

% by mass = 6.34/23.14 x 100

= 27.4%