Question

The mole fraction of potassium phosphate, K₃PO₄, in an aqueous solution is 8.11x10^-2 

The percent by mass of potassium phosphate in the solution is _______ %. 

The mole fraction of ammonium sulfide, (NH₄)₂S, in an aqueous solution is 3.42x10^-2 

The percent by mass of ammonium sulfide in the solution is _______ %.

Answer 1

1)
Consider 1 mol of solution.
Then mol of K3PO4 = 8.11*10^-2
mol of H2O = 1 - 8.11*10^-2
= 0.9189

Molar mass of H3PO4,
MM = 3*MM(H) + 1*MM(P) + 4*MM(O)
= 3*1.008 + 1*30.97 + 4*16.0
= 97.994 g/mol

use:
mass of H3PO4,
m = number of mol * molar mass
= 8.11*10^-2 mol * 97.99 g/mol
= 7.947 g

Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol

use:
mass of H2O,
m = number of mol * molar mass
= 0.9189 mol * 18.02 g/mol
= 16.55 g

use:
mass % of K3PO4 = mass of K3PO4 * 100 / mass of solution
= 7.947 * 100 / (7.947 + 16.55)
= 32.4 %

Answer: 32.4 %

2)
Consider 1 mol of solution.
Then mol of (NH4)2S = 3.42*10^-2
mol of H2O = 1 - 3.42*10^-2
= 0.9658 mol

Molar mass of (NH4)2S,
MM = 2*MM(N) + 8*MM(H) + 1*MM(S)
= 2*14.01 + 8*1.008 + 1*32.07
= 68.154 g/mol

use:
mass of (NH4)2S,
m = number of mol * molar mass
= 3.42*10^-2 mol * 68.15 g/mol
= 2.331 g

Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol

use:
mass of H2O,
m = number of mol * molar mass
= 0.9658 mol * 18.02 g/mol
= 17.4 g

use:
mass % of (NH4)2S = mass of (NH4)2S * 100 / mass of solution
= 2.331 * 100 / (2.331 + 17.4)
= 11.8 %

Answer: 11.8 %