a
We known that no. of moles \(=\text { Vlitre } \times \text { Molarity and No. of millimoles }=V_{m l} \times \text { Molarity }\)

\(\text { so millimoles of } {NaOH}=250 \times 0.5=125\)

\(\text { Millimoles of } {HCl}=500 \times 1=500\)

\(\text { Now reaction is }\)

\({NaOH}+{HCL} \rightarrow {NaCl}+{H}_{2} {O}\)

\({t}=0 \quad 125\quad 500 \quad 0 \quad 0\)

\({t}=0 \quad 0 375 \quad 125 \quad 125\)

\(\text { so millimoles of } {HCl} \text { left }=375\)

\(\text { Moles of } {HCl}=375 \times 10^{-3}\)

\(\text { No. of } {HCl} \text { molecules }=6.022 \times 10^{23} \times 375 \times 10^{-3}\)

\(=225.8 \times 10^{21}\)

\(\approx 226 \times 10^{21}=226\)