Answer is:
3,94·10⁻² moles of silver chloride are contained in 244 ml of solution.
V(AgCl) =
244 ml = 0,244 l.
b(AgCl) =
0,135 mol/kg.
d(AgCl) =
1,22 g/ml.
m(AgCl) =
V(AgCl) · d(AgCl).
m(AgCl) = 244 ml · 1,22 g/ml.
m(solution AgCl) =
297,68 g.
m(AgCl) = 0,135 mol ·143,3 g/mol = 19,34 g.
ω = 19,34 g ÷ 19,34 g + 1000 g = 0,0189.
m(AgCl) = 0,0189 · 297,68 g = 5,64 g.
n(AgCl) = 5,64 g ÷ 143,3 g/mol = 0,0394 mol.
