Determine the mass of 40.6% Na3PO4 solution that contains 6.51 g of Na+
Molar mass of Na = 22.99 g/mol
mass(Na)= 6.51 g
use:
number of mol of Na,
n = mass of Na/molar mass of Na
=(6.51 g)/(22.99 g/mol)
= 0.2832 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
= 0.2832 / 3
= 9.439*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
= 9.439*10^-2 mol * 1.639*10^2 g/mol
= 15.47 g
Now use:
mass % of Na3PO4 = mass of Na3PO4 * 100 / mass of solution
40.6 = 15.47 * 100 / mass of solution
mass of solution = 38.1 g
Answer: 38.1 g
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