a
We known that no. of moles $=\text { Vlitre } \times \text { Molarity and No. of millimoles }=V_{m l} \times \text { Molarity }$
$\text { so millimoles of } \mathrm{NaOH}=250 \times 0.5=125$
$\text { Millimoles of } \mathrm{HCl}=500 \times 1=500$
$\text { Now reaction is }$
$\mathrm{NaOH}+\mathrm{HCL} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$
$\mathrm{t}=0 \quad 125\quad 500 \quad 0 \quad 0$
$\mathrm{t}=0 \quad 0 375 \quad 125 \quad 125$
$\text { so millimoles of } \mathrm{HCl} \text { left }=375$
$\text { Moles of } \mathrm{HCl}=375 \times 10^{-3}$
$\text { No. of } \mathrm{HCl} \text { molecules }=6.022 \times 10^{23} \times 375 \times 10^{-3}$
$=225.8 \times 10^{21}$
$\approx 226 \times 10^{21}=226$