Calculate the molarity of Na+ in 135 mL of a solution with 5.32 g NaCl and 1.20 g Na2SO4. Please tell and/or show how you got the answer.?

Molar mass of NaCl = (23.0 + 35.5) g/mol = 58.5 g/mol
Molar mass of Na₂SO₄ = (23.0×2 + 32.1 + 16.0×4) g/mol = 142.1 g/mol

1 mole of NaCl contains 1 mole of Na⁺.
No. of moles of Na⁺ in NaCl = [(5.32 g) / (58.5 g/mol)]

1 mole of Na₂SO₄ contains 2 moles of Na⁺.
No. of moles of Na⁺ in Na₂SO₄ = [(1.20 g) / (142.1 g/mol)] × 2

Total number of moles of Na⁺ = [(5.32 g) / (58.5 g/mol)] + [(1.20 g) / (142.1 g/mol)] × 2 = 0.108 mol
Volume of the solution = 135 mL = 0.135 L

Molarity of Na⁺ = (0.108 mol) / (0.135 L) = 0.8 mol/L = 0.8 M

Molar mass of NaCl is 58.4428 g/mol
Mol NaCl in 5.32g = 5.32g/58.4428g/mol = 0.0910 mol NaCl
NaCl dissociates:
NaCl(aq) → Na+(aq) + Cl- (aq)
1mol NaCl produces 1 mol Na+ ions
The solution contains 0.0910 mol Na+ ions from the NaCl

Molar mass of Na2SO4 is 142.04 g/mol
Mol Na2SO4 in 1.20g = 1.20g/142.04g/mol = 0.00845 mol Na2SO4
Na2SO4 dissociates:
Na2SO4(aq) → 2Na+(aq) + SO4 2-(aq)
1mol Na2SO4 produces 2 mol Na+ ions
The solution contains 0.00845*2 = 0.0169 mol Na+ ions from the Na2SO4

Total mol Na+ ions = 0.0910mol + 0.0169 mol = 0.1079 mol Na+ ions
Volume of solution = 135mL = 0.135L
Molarity of Na+ ions = 0.1079mol / 0.135L
Molarity of Na+ ions = 0.799M