250   mL of 0 . 5 M   NaOH was added to 500   mL of 1 M HCl . The number of unreacted HCl…

We known that no. of moles $={\mathrm{V}}_{\text{litre }}\times$ Molarity $\&$ No. of millimoles $={\mathrm{V}}_{\mathrm{ml}}\times$ Molarity so millimoles of $\mathrm{NaOH}=250\times 0.5$
$=125$
Millimoles of $\mathrm{HCl}=500\times 1=500$
Now reaction is

so millimoles of $\mathrm{HCl}$ left $=375$ Moles of $\mathrm{HCl}=375\times {10}^{-3}$
No. of $\mathrm{HCl}$ molecules $=6.022\times {10}^{23}\times 375\times {10}^{-3}$
$=225.8\times {10}^{21}$
$\simeq 226\times {10}^{21}=226$