Question

Determine the mass of 2.74% AlF3 solution that contains 4.37 g of F-

Answer 1

Molar mass of F = 19 g/mol


mass(F)= 4.37 g

use:
number of mol of F,
n = mass of F/molar mass of F
=(4.37 g)/(19 g/mol)
= 0.23 mol
This is number of moles of F

one mole of AlF3 contains 3 mole of F
use:
number of moles of AlF3 = number of moles of AlF3 / 3
= 0.23 / 3
= 7.667*10^-2

Molar mass of AlF3,
MM = 1*MM(Al) + 3*MM(F)
= 1*26.98 + 3*19.0
= 83.98 g/mol

use:
mass of AlF3,
m = number of mol * molar mass
= 7.667*10^-2 mol * 83.98 g/mol
= 6.438 g

Now use:
mass % of AlF3 = mass of AlF3 * 100 / mass of solution
2.74 = 6.438 * 100 / mass of solution
mass of solution = 235 g
Answer: 235 g