1. Write the reaction

$\mathrm{NaOH + HCl \rightarrow NaCl + H_2O}$

This is a $1:1$ neutralization reaction.

2. Calculate moles of NaOH

Volume of NaOH $= 250\,\text{mL} = 0.250\,\text{L}$

Molarity of NaOH $= 0.5\,\text{M}$

$n(\mathrm{NaOH}) = M \times V = 0.5 \times 0.250 = 0.125\,\text{mol}$

3. Calculate moles of HCl

Volume of HCl $= 500\,\text{mL} = 0.500\,\text{L}$

Molarity of HCl $= 1\,\text{M}$

$n(\mathrm{HCl}) = 1 \times 0.500 = 0.500\,\text{mol}$

4. Find the limiting reagent and unreacted HCl

Since the reaction ratio is $1:1$, $0.125$ mol NaOH will react with $0.125$ mol HCl.

Initial HCl moles $= 0.500$ mol

Unreacted HCl moles:

$0.500 - 0.125 = 0.375\,\text{mol}$

5. Convert unreacted HCl moles to number of molecules

$N = 0.375 \times 6.022 \times 10^{23}$

$N = 2.25825 \times 10^{23}$

Now express this as $x \times 10^{21}$:

$2.25825 \times 10^{23} = 225.825 \times 10^{21}$

Nearest integer:

$x = 226$

6. Final answer

The number of unreacted HCl molecules is:

$\boxed{226 \times 10^{21}}$

So the required integer is:

$\boxed{226}$

7. Comparison with stored answer

Stored correct answer $= 226$

Our derived answer also $= 226$, so they agree.