Question

If you have an aqueous solution that is 37.2 % Na3PO4 by mass, what is the molality of Na3PO4 in the solution?

Enter your answer in units of molality to three significant figures.

Answer 1

Let mass of solution be 1 Kg = 1000 g

mass of Na3PO4 = 37.2 % of mass of solution

= 37.2*1000.0/100

= 372.0 g

mass of solvent = mass of solution - mass of solute

mass of solvent = 1000 g - 372.0 g

mass of solvent = 628.0 g

mass of solvent = 0.628 Kg

Molar mass of Na3PO4,

MM = 3*MM(Na) + 1*MM(P) + 4*MM(O)

= 3*22.99 + 1*30.97 + 4*16.0

= 163.94 g/mol

mass(Na3PO4)= 372.0 g

use:

number of mol of Na3PO4,

n = mass of Na3PO4/molar mass of Na3PO4

=(3.72*10^2 g)/(1.639*10^2 g/mol)

= 2.269 mol

m(solvent)= 0.628 Kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(2.269 mol)/(0.628 Kg)

= 3.613 molal

Answer: 3.61 molal