48. Determine the molarity of each of the following solutions:

(a) 1.457 mol KCl in 1.500 L of solution

(b) 0.515 g of H₂SO₄ in 1.00 L of solution

(c) 20.54 g of Al(NO₃)₃ in 1575 mL of solution

(d) 2.76 kg of CuSO₄∙5H₂O in 1.45 L of solution

(e) 0.005653 mol of Br₂ in 10.00 mL of solution

(f) 0.000889 g of glycine, C₂H₅NO₂, in 1.05 mL of solution

1.457 mol KCl in 1.500 L of solution

Mole of KCl = 1.457

Molarity = Moles of solute / Liters of solution.

Molarity = 1.457 mol/ 1.5 L = 0.971 M

0.515 g of H₂SO₄ in 1.00 L of solution

Mass of H₂SO₄ = 0.515 g

Molecular weight of H₂SO₄ = 2x 1 + 32 + 4 x 16 = 98 g/mol

Moles of H₂SO₄ = 0.515 g / 98 g/mol = 0.0052 Moles

Molarity = Moles of solute/ Liters of solution

= 0.0052 Moles/ 1 L

= 0.0052 M

20.54 g of Al(NO₃)₃ in 1575 mL of solution

1575 mL = 1.575 L ( 1 L = 1000 mL)

Mass of Al(NO₃)₃ = 20.54 g

Molecular weight of Al(NO₃)₃ = 30 + 3 x 14 + 9 x 16 = 30 + 42+ 144 = 216 g/mol

Moles of Al(NO₃)₃ = 20.54 g/ 216 g/mol = 0.0950 mole

Molarity = 0.0950 mole/ 1.575 L = 0.0603 M

2.76 kg of CuSO₄∙5H₂O in 1.45 L of solution

Mass ofCuSO₄∙5H₂O = 2.76 kg = 2760 g (1kg = 1000 g)

Molecular weight of CuSO₄∙5H₂O = 63 + 32 + 9 x 16 + 10 x 1 = 95 + 154 = 249 g/mol

Moles of CuSO₄∙5H₂O = 2760 g / 249 g/mol = 11.084 mole

Molarity of CuSO₄∙5H₂O = 11.084 mole/ 1.45 L

= 7.644 M

0.005653 mol of Br₂ in 10.00 mL of solution

10 mL = 0.01L ( 1L = 1000 mL )

Molarity of Br₂ = Moles of solute / Liters of solution

Molarity of Br₂ = 0.005653 mol/ 0.1 L

= 0.05653 M

0.000889 g of glycine, C₂H₅NO₂, in 1.05 mL of solution

1.05 mL = 0.00105 L ( 1 L = 1000 mL )

Molecular weight of C₂H₅NO₂ = 2x 12 + 5 x 1+ 14 + 2 x 16 = 29 + 14 + 32 = 75 g/mol

Moles of C₂H₅NO₂ = 0.000889 g / 75 g/mol = 1.185 x 10-5 mole

Molarity = 1.185 x 10-5 mole/ 0.00105 L

= 0.0112 M