Question

A sample of a metal chloride, MCl 2, dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed weighed 3.60 Calculate the molar mass of M.

Answer 1

$$n(AgCl) = m/M = 3.60/143.32 = 0.025 mol$$ $$n(MCl_2) = n(AgCl)/2 = 0.0125 mol$$ $$n(Cl) = 0.025 mol$$ $$m(Cl) = 0.025*35.5 = 0.08875g$$