If all of the chloride in a 3.734-g sample of an unknown metal chloride is precipitated as agcl with 70.90 ml of 0.2010 m agno3, what is the percentage of chloride in the sample?

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 %.

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .

the answer would be a