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)

Solution

Chemistry ma’am · 4 steps

Track the moles of acid, not the concentration: subtract what boiled off, then divide by the new volume.

1. Moles at start. Moles = molarity × volume in litres = 0.5 × 0.8 = 0.4 mol of HNO₃.
2. Moles lost. The evaporated acid is 11.5/63 = 0.1825 mol. So the moles left in the beaker are 0.4 - 0.1825 = 0.2175 mol.
3. New volume. The volume falls to half, so it is 400 mL = 0.4 L.

   M = (0.2175 mol)/(0.4 L) = 0.5437 M = 54.37 × 10⁻² M

4. So. x = 54.37, which rounds to 54.

The common slip: 27. Dividing the remaining 0.2175 mol by the original 0.8 L instead of the halved 0.4 L gives 27.

The idea to keep. When a solution is heated or diluted, the moles of solute change only if solute actually leaves. Count the moles that remain first, then divide by the final volume.

Answer from the NTA final answer key. Solution by PrepWiser.