Question

a) In the laboratory you dissolve 12.4 g of lead nitrate in a volumetric flask and add water to a total volume of 500 . mL.

What is the molarity of the solution? _____M

b)How many grams of Cu(OH)₂ will precipitate when excess KOH solution is added to 43.0 mL of 0.579 M CuSO₄ solution?

CuSO₄(aq) + 2KOH(aq) Cu(OH)₂(s) + K₂SO₄(aq)

Answer 1

To find the molarity of the solution Given that 12.4g of lead nitrate 500.0 mL of solution First find the moles of lead nitrate(Pb(NO3)) mass moles =- molarm ass molarmass of lead nitrate (Pb(NO3)2) = 3312g/mol 12.4g moles = 5 331.2g/mol = 0.0374moles To find the molarity moles of solute Molarity = liters of solution 500.0mL = 0.5L 0.0374moles 0.5L = 0.0748M Hence the molarity of the solution = 0.0748M

(6) To find the grames of Cu(OH), Given that 43.0mL of 0.579M CuSO, solution CuSO4 (aq) + 2OH(aq) → Cu(OH),(s) + K 50, (aq) First find the moles of Cuso, moles = molarity X volume 43.Om L = 0.043L moles = molarity x volume = 0.579mol/L 0.043L = 0.0248moles To find the moles of Cu(OH), According to reactio Cu(OH), and Cuso, is 1:1 mole ratio Therefore moles of Cu(OH), = 0.0248m oles To find the mass of Cu(OH), mass = moles Xmolarm ass molarmass of Cu(OH), = 97.561g/mol mass = 0.0248mol 97.561g/mol = 2.41g Hence the mass of Cu(OH), = 2.41g|