Answer is: the percentage of chloride in the sample is 13.53%.
Balanced chemical reaction: MCl + AgNO₃(aq) → AgCl(s) + MNO₃(aq).
V(AgNO₃) = 70.90 mL ÷ 1000 mL.
V(AgNO₃) = 0.0709 L; volume of silver nitrate.
c(AgNO₃) = 0.2010 M; molarity of the solution.
n(AgNO₃) = V(AgNO₃) · c(AgNO₃).
n(AgNO₃) = 0.0709 L · 0.201 mol/L.
n(AgNO₃) = 0.0143 mol; amount of silver nitrate.
From chemical reaction: n(AgNO₃) : n(MCl) = 1 : 1.
n(MCl) = 0.0143 mol; amount of the sample.
M(MCl) = m(MCl) ÷ n(MCl).
M(MCl) = 3.734 g ÷ 0.0143 mol.
M(MCl) = 261.11 g/mol; molar mass of the sample.
ω(Cl) = M(Cl) ÷ M(MCl) · 100%.
ω(Cl) = 35.45 g/mol ÷ 261.11 g/mol · 100%.
ω(Cl) = 13.53 %.
