Answer to Question #131291 in Chemistry for Kate

Question

Calculate the normality of the following:
a. 26.5 grams nitric acid in 2L of solution
b. 150g of calcium carbonate in 500 ml solution

Answer 1

Solution:

Normality (N) It is the number of gram equivalent of solute present in one litre of solution.

• Normality = Number of gram equivalents / Volume of solution in litres
• Number of gram equivalents = Weight of solute / Equivalent weight of solute
• N = Weight of Solute (gram) / [Equivalent weight × Volume (L)]

Part (a):

nitric acid = HNO₃

To find Equivalent weight of HNO₃, use the formula E_q = MW / n

• E_q = Equivalent weight
• MW = Molecular weight in grams/mole
• n = number of equivalents

For every mole of nitric acid, there is one H⁺ ions, n = 1.

Molecular weight (MW) of HNO₃ is 63.01 g/mol.

Consequently,

E_q = (63.01 g/mol) / 1 = 63.01 g/mol

Thus:

Normality of HNO₃ = (26.5 g) / [63.01 g/mol × 2 L] = 0.21 N

Normality of HNO₃ is 0.21 N

Answer (a): Normality of HNO₃ is 0.21 N.

Part (b):

calcium carbonate = CaCO₃

To find Equivalent weight of CaCO₃, use the formula E_q = MW / n

Molecular weight (MW) of CaCO₃ is 100.087 g/mol.

Valency (Ca²⁺; CO₃^2-) is the n-factor for salt, n = 2.

Consequently,

E_q = (100.087 g/mol) / 2 = 50.0435 g/mol

Thus:

Normality of CaCO₃ = (150 g) / [50.0435 g/mol × 0.5 L] = 5.995 N = 6N

Normality of CaCO₃ is 6 N

Answer (b): Normality of CaCO₃ is 6 N.