The chloride is precipitated as following:
AgNO₃ + Cl⁻ → AgCl + NO₃⁻
molar concentration = number of moles / solution volume
number of moles = molar concentration × solution volume
number of moles of AgNO₃ = 0.2010 × 70.90
number of moles of AgNO₃ = 14.25 mmoles
from the equation
if 1 mmole of AgNO₃ is needed to precipitate 1 mmole of chloride
then 14.25 mmoles of AgNO₃ are needed to precipitate X mmoles of
chloride
X = 14.25 mmoles of choride
mass = number of moles × molecular weight
mass of chloride = 14.25 × 35.5
mass of chloride = 505.9 mg = 0.5059 g
if in 4.106 g of sample we have 0.5059 g of chloride
then in 100 g of sample we have X g of chloride
X = (100 × 0.5059) / 4.106 = 12.32 % chloride
