250 mL of 0.5M NaOH was added to 500 mL of 1M HCl. The number of unreacted HCl molecules in the solution is p$$\times 10^{21}$$. Find out p.
(Nearest integer) (N$$_A$$ = 6.022 $$\times 10^{23}$$)

First, find the moles of NaOH and HCl present:

Moles of NaOH = $$0.5 \text{ M} \times 0.250 \text{ L} = 0.125 \text{ mol}$$

Moles of HCl = $$1.0 \text{ M} \times 0.500 \text{ L} = 0.500 \text{ mol}$$

The neutralization reaction is: NaOH + HCl → NaCl + H₂O. NaOH is the limiting reagent, so 0.125 mol of HCl is consumed.

Moles of unreacted HCl = $$0.500 - 0.125 = 0.375 \text{ mol}$$

Number of HCl molecules = $$0.375 \times 6.022 \times 10^{23} = 2.258 \times 10^{23}$$

Expressing as $$p \times 10^{21}$$: $$p = \frac{2.258 \times 10^{23}}{10^{21}} = 225.8 \approx 226$$

Therefore, $$p = \boxed{226}$$.