Question

  

When a solution containing silver ions is mixed with another solution containing chloride ions, a precipitate of silver chloride forms. When 85.00 ml of a silver nitrate solution is mixed with an excess of a sodium chloride solution, all of the silver ion is precipitated as silver chloride. The solid is collected, washed, dried, and found to have a mass of 6.5314 g. Calculate the molarity of the original silver nitrate solution.

Answer 1

Molar mass of AgCl = 143.3 g/mol

Mass of AgCl obtained = 6.5314 g.

Amount of AgCl = Mass / molar mass = 6.5314 / 143.3 = 0.04558 mol

AgNO₃ (aq.) + NaCl (aq.) -----------> AgCl (s) + NaNO₃ (aq.)

Fromt the balanced eqution,

1 mol AgCl is fromed from 1 mol AgNO₃

Then, 0.04558 mol AgCl is fromed from 0.04558 mol AgNO₃

Therefore,

Amount of AgNO3 = 0.04558 mol

Volume of AgNO3 solution = 85.00 mL = 0.08500 L

Now,

Molarity of AgNO₃ = amount of AgNO₃ / Volume of solution in L

M = 0.04558 / 0.08500

M = 0.5362 M