A weighed portion of 161 g of Glauber’s salt Na2CO3 * 10H2O was dissolved in 180 l of water.

A weighed portion of 161 g of Glauber’s salt Na2CO3 * 10H2O was dissolved in 180 l of water. Calculate the mass fraction of sodium sulfate in the resulting solution. How many ions of each type are in it?

Given:
m (salt) = 161 g.
V (H2O) = 180 L = 180,000 ml (m = 180,000 g)
w (Na2SO4) -?

Decision:
1) the mass of the solution after dissolution is 161 + 180000 = 180161 g
2) 322 g of salt – 142 g of Na2SO4 (based on M)
161 g salt = x g Na2SO4
x = 142 * 161/322 = 71 g.
3) w (Na2SO4) = 71/180161 = 0.00039 = 0.039%
4) n (Na2SO4) = 71/142 = 0.5 mol
5) n (Na) = 2n (Na2SO4) = 1 mol
6) n (SO4) = n (Na2SO4) = 0.5 mol

Answer: 0.039%, 0.5 mol, 1 mol.