The volume of HCl, containing 73 g L⁻¹, required to completely neutralise NaOH obtained by reacting 0.69 g of metallic sodium with water, is __________ mL. (Nearest Integer)
(Given : molar masses of Na, Cl, O, H, are 23, 35.5, 16 and 1 g mol⁻¹ respectively.)

Solution

Chemistry ma’am · 4 steps

Follow the moles: sodium gives the same number of moles of NaOH, and that sets the HCl needed.

1. Moles of sodium. n(Na) = 0.69/23 = 0.03 mol. In 2Na + 2H₂O → 2NaOH + H₂, each Na gives one NaOH, so n(NaOH) = 0.03 mol.
2. Molarity of HCl. The molar mass of HCl is 1 + 35.5 = 36.5 g mol⁻¹. So 73 g L⁻¹ is 73/36.5 = 2 mol L⁻¹.
3. Neutralisation. NaOH + HCl → NaCl + H₂O is 1 : 1, so you need 0.03 mol of HCl.

   V = n/M = (0.03 mol)/(2 mol L⁻¹) = 0.015 L

4. So. 0.015 L = 15 mL, so the answer is 15.

The common slip: 7.5. Reading the 2 : 2 equation as if 2 Na give only 1 NaOH halves the moles to 0.015 mol and gives 7.5 mL.

The idea to keep. In a chain of reactions, carry moles from one step to the next using the coefficients. Turn a g per litre concentration into molarity before you find a volume.

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