What is the normality for a solution containing 100 g of NaCl made up to 500 mL with distilled water? Assume a gram molecular weight of approximately 58 g/mol.

What is the normality for a solution containing 100 g of NaCl made up to 500 mL with distilled water? Assume a gram molecular weight of approximately 58 g/mol.

• A. 3.45
• B. 0.86
• C. 1.72
• D. 6.9

Answer: (A)

Normality is the concentration of reactive equivalents per liter. For NaCl, the equivalent factor is 1, since dissociation into Na+ and Cl− provides one reactive unit per mole in typical acid–base or redox contexts. So normality equals molarity. First find the amount of NaCl in moles: 100 g / 58 g/mol ≈ 1.72 mol. The solution volume is 0.50 L, so the molarity is 1.72 mol / 0.50 L ≈ 3.44 M. With a valence factor of 1, the normality is the same: about 3.44 N, which rounds to 3.45 N. Thus the best choice is 3.45. The other numbers would require a different interpretation of the equivalent factor or a bookkeeping error, but for NaCl the normality matches the molarity.

Normality is the concentration of reactive equivalents per liter. For NaCl, the equivalent factor is 1, since dissociation into Na+ and Cl− provides one reactive unit per mole in typical acid–base or redox contexts. So normality equals molarity.

First find the amount of NaCl in moles: 100 g / 58 g/mol ≈ 1.72 mol. The solution volume is 0.50 L, so the molarity is 1.72 mol / 0.50 L ≈ 3.44 M. With a valence factor of 1, the normality is the same: about 3.44 N, which rounds to 3.45 N.

Thus the best choice is 3.45. The other numbers would require a different interpretation of the equivalent factor or a bookkeeping error, but for NaCl the normality matches the molarity.