250mL of 0 . 5MNaOH was added to 500mL of 1M HCl . The number of...

We known that no. of moles $= {V}_{litre } \times$ Molarity & No. of millimoles $= {V}_{ml} \times$ Molarity so millimoles of $NaOH = 250 \times 0 . 5$
$= 125$
Millimoles of $HCl = 500 \times 1 = 500$
Now reaction is

so millimoles of $HCl$ left $= 375$~Moles of $HCl = 375 \times {10}^{- 3}$
No. of $HCl$ molecules $= 6 . 022 \times {10}^{23} \times 375 \times {10}^{- 3}$
$= 225 . 8 \times {10}^{21}$
$≃ 226 \times {10}^{21} = 226$