Question

When 800 mL of 0.5 M nitric acid is heated in a beaker, its volume is reduced to half and 11.5 g of nitric acid is evaporated. The molarity of the remaining nitric acid solution is \(x × 10^{–2} M\). (Nearest integer)
(Molar mass of nitric acid is 63 g mol^–1)

Answer 1

Correct Answer: 54

To calculate the molarity of the remaining nitric acid solution after heating, we follow these steps:

1. Initial Amount of HNO₃:
   • Initial volume of solution = 800 mL = 0.8 L.
   • Given molarity = 0.5 M, thus initial moles of HNO₃ = 0.5 mol/L × 0.8 L = 0.4 mol.
2. HNO₃ Evaporated:
   • Molar mass of HNO₃ = 63 g/mol.
   • Moles of HNO₃ evaporated: 11.5 g ÷ 63 g/mol = 0.1825 mol.
3. Remaining Moles of HNO₃:
   • Remaining moles = 0.4 mol - 0.1825 mol = 0.2175 mol.
4. Final Volume of Solution:
   • Since the volume is reduced to half: Final volume = 0.8 L ÷ 2 = 0.4 L.
5. Molarity of Remaining Solution:
   • Molarity = moles/volume = 0.2175 mol ÷ 0.4 L = 0.54375 M.
6. Convert to Required Format:
   • Molarity expressed as \(x × 10^{–2}\, M\):
     x = 54.375 ≈ 54 (nearest integer).
7. Validation:
   • The value x = 54 fits the range 54,54.

The molarity of the remaining nitric acid solution is \(54 × 10^{-2} M\).