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$1 \mathrm{~mL}$ of water has 25 drops. Let $\mathrm{N}_0$ be the Avogadro number. What is the number of molecules present in 1 drop of water ? (Density of water $=1 \mathrm{~g} / \mathrm{mL}$ )
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The correct answer is:
$\frac{0.02}{9} \mathrm{~N}_0$
Volume of one drop $=\left(\frac{1}{25}\right) \mathrm{mL}$
$\therefore$ Mass of $1 \mathrm{drop}=\mathrm{V} \times \mathrm{d}$
$=\left(\frac{1}{25} \mathrm{~mL}\right)(1 \mathrm{~g} / \mathrm{mL})$
$=\frac{1}{25} \mathrm{~g}$
Number of moles of $\mathrm{H}_2 \mathrm{O}=\frac{\text { Massof water inonedrop }}{\text { Molarmassof water }}=\frac{\frac{1}{25}}{18}=\frac{1}{25 \times 18}$
$\therefore$ Number of $\mathrm{H}_2 \mathrm{O}$ Molecule $=\frac{1}{25 \times 18} \mathrm{~N}_0=\frac{1}{50 \times 9} \mathrm{~N}_0=\frac{0.02}{9} \mathrm{~N}_0$
$\therefore$ Mass of $1 \mathrm{drop}=\mathrm{V} \times \mathrm{d}$
$=\left(\frac{1}{25} \mathrm{~mL}\right)(1 \mathrm{~g} / \mathrm{mL})$
$=\frac{1}{25} \mathrm{~g}$
Number of moles of $\mathrm{H}_2 \mathrm{O}=\frac{\text { Massof water inonedrop }}{\text { Molarmassof water }}=\frac{\frac{1}{25}}{18}=\frac{1}{25 \times 18}$
$\therefore$ Number of $\mathrm{H}_2 \mathrm{O}$ Molecule $=\frac{1}{25 \times 18} \mathrm{~N}_0=\frac{1}{50 \times 9} \mathrm{~N}_0=\frac{0.02}{9} \mathrm{~N}_0$
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