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)

We need to find the volume of HCl solution required to completely neutralize the NaOH obtained from the reaction of metallic sodium with water.

Find moles of Na.

Molar mass of Na = 23 g/mol

$$ \text{Moles of Na} = \frac{0.69}{23} = 0.03 \text{ mol} $$

Reaction of Na with water.

$$ 2\text{Na} + 2\text{H}_2\text{O} \to 2\text{NaOH} + \text{H}_2 $$

Moles of NaOH formed = Moles of Na = 0.03 mol (1:1 ratio)

Neutralization reaction.

$$ \text{NaOH} + \text{HCl} \to \text{NaCl} + \text{H}_2\text{O} $$

Moles of HCl required = Moles of NaOH = 0.03 mol

Find the molarity of HCl solution.

HCl solution contains 73 g/L.

Molar mass of HCl = $$1 + 35.5 = 36.5$$ g/mol

$$ \text{Molarity} = \frac{73}{36.5} = 2 \text{ M} $$

Calculate the required volume.

$$ V = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.03}{2} = 0.015 \text{ L} = 15 \text{ mL} $$

The volume of HCl required is $$\boxed{15}$$ mL.