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Question

Calculate the number of water molecules present in one drop of water which weighs $0.18 g$ .

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Solution

Molecular mass of $H_{O} = 18 g = 1 m o l e$
$18 g$ of water contains $6.022 \times 10^{23}$ molecules.
$∴ 0.18 g$ of water contains, $= \frac{0.18 \times 6.023 \times 10^{23}}{18} = 0.06023 \times 10^{23} = 6.023 \times 10^{23} molecules$ .

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