Question

Determine the mass of 21.6% Na3PO4 solution that contains 4.61 g of Na+.

Answer 1

use:

number of molecules = number of mol * Avogadro’s number

number of molecules = 2.73 * 6.022*10^23 molecules

number of molecules = 1.644*10^24 molecules

Molar mass of Na = 22.99 g/mol

mass(Na)= 4.61 g

use:

number of mol of Na,

n = mass of Na/molar mass of Na

=(4.61 g)/(22.99 g/mol)

= 0.2005 mol

This is number of moles of Na

one mole of Na3PO4 contains 3 mole of Na

use:

number of moles of Na3PO4 = number of moles of Na3PO4 / 3

= 0.2005 / 3

= 6.684*10^-2

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

use:

mass of Na3PO4,

m = number of mol * molar mass

= 6.684*10^-2 mol * 1.639*10^2 g/mol

= 10.96 g

Now use:

Mass % of Na3PO4 = mass of Na3PO4 * 100 / mass of solution

21.6 = 10.96 * 100 / mass of solution

mass of solution = 50.7 %

Answer: 50.7 %