If all of the chloride in a 4.106-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?

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