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The total number of electrons in $18 \mathrm{~mL}$ of water (density $=1 \mathrm{~g} \mathrm{~mL}^{-1}$ ) is
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Verified Answer
The correct answer is:
$6.02 \times 10^{24}$
In $18 \mathrm{~mL}$, number of moles of
$\mathrm{H}_{2} \mathrm{O}=\frac{\text { mass of } \mathrm{H}_{2} \mathrm{O}}{\text { molecular mass }}$
$=\frac{\text { density } \times \text { volume }}{\text { molecular mass }}$
$=\frac{1 \times 18}{18}=1 \mathrm{~mol}$
$\because$ Number of moles of $\mathrm{H}_{2} \mathrm{O}$ in $1 \mathrm{~mol}$
$=6.022 \times 10^{23}$
and number of $\mathrm{e}^{-}$in 1 molecule of $\mathrm{H}_{2} \mathrm{O}$
$=1 \times 2+8=10$
$\therefore$ Number of $\mathrm{e}^{-}$in 1 mole of $\mathrm{H}_{2} \mathrm{O}$
$=6.022 \times 10^{23} \times 10$
$=6.022 \times 10^{24}$
$\mathrm{H}_{2} \mathrm{O}=\frac{\text { mass of } \mathrm{H}_{2} \mathrm{O}}{\text { molecular mass }}$
$=\frac{\text { density } \times \text { volume }}{\text { molecular mass }}$
$=\frac{1 \times 18}{18}=1 \mathrm{~mol}$
$\because$ Number of moles of $\mathrm{H}_{2} \mathrm{O}$ in $1 \mathrm{~mol}$
$=6.022 \times 10^{23}$
and number of $\mathrm{e}^{-}$in 1 molecule of $\mathrm{H}_{2} \mathrm{O}$
$=1 \times 2+8=10$
$\therefore$ Number of $\mathrm{e}^{-}$in 1 mole of $\mathrm{H}_{2} \mathrm{O}$
$=6.022 \times 10^{23} \times 10$
$=6.022 \times 10^{24}$
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