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

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.