Question

Calculate the molarity and normality of the following compounds:

a.80 μg/L HNO^3

b.135 μg/L CaCO^3

c.10 μg/L Cr(OH)^3 – Chromium(III) hydroxide will dissociate in water to form a chromium ion and 3 hydroxide ions. The charge of 1 hydroxide ion is -1. The number (III) should also give you a hint to determine n.

Answer 1

Molarity = concentration/Mw (mol/Li)

Normality = concentration/Equivalent weight

1. 80 μg/L HNO₃

Molecular weight = 1+14+3 x 16 = 63 g/mol

80 μg/L = 80 x 10^-6 g/L

Molarity = 80 x 10^-6 g/L ÷ 63 g/mol = 1.27 x 10^-6 mol/L = 1.27 x 10^-6 M

Equivalent weight = molecular weight/basicity = 63/1 = 63 g/mol

Normality = 80 x 10^-6 g/L ÷ 63 g/equivalent = 1.27 x 10^-6 equiv/L = 1.27 x 10^-6 N

1. 135 μg/L CaCO₃

Molecular weight = 40+12+3 x 16 = 100 g/mol

135 μg/L = 135 x 10^-6 g/L

Molarity = 135 x 10^-6 g/L ÷ 100 g/mol = 1.35 x 10^-6 mol/L = 1.35 x 10^-6 M

Equivalent weight = molecular weight/oxidation number = 100/2 = 50 g/mol

Normality = 135 x 10^-6 g/L ÷ 50 g/equivalent = 2.7 x 10^-6 equiv/L = 2.7 x 10^-6 N

1. 10 μg/L Cr(OH)₃

Molecular weight = 52+ 3 x 16 +3 = 103 g/mol

10 μg/L = 10 x 10^-6 g/L

Molarity = 10 x 10^-6 g/L ÷ 103 g/mol = 0.097 x 10^-6 mol/L = 9.7 x 10^-8 M

Equivalent weight = molecular weight/oxidation number = 103/3 = 34.3 g/mol

Normality = 10 x 10^-6 g/L ÷ 34.3 g/equivalent = 2.9 x 10^-7 equiv/L