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WBJEE 2022 Physics and Chemistry Paper

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Question : 45 of 80

Marks: +1, -0

1\ \mathrm{mL}

of water has 25 drops. Let

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})

$\frac{0.02}{9} N_{0}$
$\frac{18}{25} N_{0}$
$\frac{25}{18} N_{0}$
$\frac{0.04}{25} N_{0}$

Solution:

Hint: Volume of one drop

=\left(\frac{1}{25}\right)\ \mathrm{mL}

\therefore

Mass of 1 drop

=V \times 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 in one drop}}{\text{Molarmass of 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} N_{0} = \frac{1}{50 \times 9} N_{0} = \frac{0.02}{9} N_{0}

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