Question

1. Show the calculation of the mass of Ca3(PO4)2 needed to make 200 ml of a 0.128 M solution.

2. Show the calculation of the volume of 0.987 M solution which can be prepared using 24.6 grams of NaNO3.

3. Show the calculation of the volume of 0.238 M solution which can be prepared using 13.4 grams of Ca3(PO4)2.

Answer 1

(1) Given molarity of Ca3(PO4)2=0.128 M=0128 mol/L,

Volume=200 mL=0.2 L.

Moles =molarity x volume=0.128 mol/L x 0.2 L=0.0256 mol

Molar mass of Ca3(PO4)2=310.18 g/mol

Mass of Ca3(PO4)2=moles x molar mass=0.0256 mol x 310 18 g/mol=7.94 g.

(2) Given mass of NaNO3=24.6 g and molar mass of NaNO3=84.9947 g/mol.

Therefore moles of NaNO3=mass/molar mass=24.6 g/84.9947 g/mol=0.2894 mol.

Molarity of NaNO3=0.987 M=0.987 mol/L

Molarity=moles / volume

Volume=moles/molarity=0.2894 mol/0.987 mol/L=0.293 L

Volume=293 mL.

(3) Given molarity of Ca3(PO4)2=0.238 M=0.238 mol/L

Mass=13.4 g and molar mass=310.18 g/mol.

Moles of Ca3(PO4)2=13.4 g/310.18 g/mol=0.0432 mol.

Therefore volume=moles/molarity=0.0432 mol/0.238 mol/L=0.1815 L=181.5 mL.

Please let me know if you have any doubt. Thanks.