250 mL of 0.5 M NaOH was added


$250 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{NaOH}$ was added to $500 \mathrm{~mL}$ of $1 \mathrm{M}$ $\mathrm{HCl}$. The number of unreacted $\mathrm{HCl}$ molecules in the solution after complete reaction is $\times 10^{21}$. (Nearest integer) 

$\left(\mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23}\right)$


We known that no. of moles $=\mathrm{V}_{\text {litre }} \times$ Molarity

\& No. of millimoles $=\mathrm{V}_{\mathrm{ml}} \times$ Molarity

so millimoles of $\mathrm{NaOH}=250 \times 0.5$


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 HCl molecules $=6.022 \times 10^{23} \times 375 \times 10^{-3}$

$=225.8 \times 10^{21}$

$\simeq 226 \times 10^{21}=226$

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