Question

An aqueous solution is made by dissolving 21.2 grams of ammonium phosphate in 323 grams of water.

The molality of ammonium phosphate in the solution is ? m.

In the laboratory you are asked to make a 0.150 m potassium sulfide solution using 15.7 grams of potassium sulfide.  

How much water should you add?  
grams

If 20.9 grams of an aqueous solution of ammonium phosphate, (NH₄)₃PO₄, contains 3.33 grams of ammonium phosphate, what is the percentage by mass of ammonium phosphate in the solution?

% (NH₄)₃PO₄

How many grams of KF are there in 195 grams of an aqueous solution that is 10.8 % by weight KF.

g KF

Answer 1

1)

Molar mass of (NH4)3PO4,

MM = 3*MM(N) + 12*MM(H) + 1*MM(P) + 4*MM(O)

= 3*14.01 + 12*1.008 + 1*30.97 + 4*16.0

= 149.096 g/mol

mass((NH4)3PO4)= 21.2 g

use:

number of mol of (NH4)3PO4,

n = mass of (NH4)3PO4/molar mass of (NH4)3PO4

=(21.2 g)/(1.491*10^2 g/mol)

= 0.1422 mol

m(solvent)= 323 g

= 0.323 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(0.1422 mol)/(0.323 Kg)

= 0.4402 molal

Answer: 0.440 molal

2)

Molar mass of K2S,

MM = 2*MM(K) + 1*MM(S)

= 2*39.1 + 1*32.07

= 110.27 g/mol

mass(K2S)= 15.7 g

use:

number of mol of K2S,

n = mass of K2S/molar mass of K2S

=(15.7 g)/(1.103*10^2 g/mol)

= 0.1424 mol

use:

molality = number of mol / mass of solvent in Kg

0.15 = 0.1424/ mass of solvent

mass of solvent = 0.949 Kg

= 949 g

Answer: 949 g

3)

mass % = mass of (NH4)3PO4 * 100 / mass of solution

= 3.33*100/20.9

= 15.9 %

Answer: 15.9 %

4)

mass % = mass of KF * 100 / mass of solution

10.8 = mass of KF * 100 / 195

mass of KF = 21.06 %

Answer: 21.1 %