What is the mass of a water molecules in gram? How many molecules are present in one drop of pure water which weighs $0.05 g$ ? If the same drop of water evaporates in one hour, calculate the number of molecules leaving the liquid surface per second.

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Solution

$Mass of water molecule = \frac{18 g}{N_{A}} = \frac{18}{6.023 \times 10^{23}} = 2.988 \times 10^{- 23} g In 18 gm there are 6.023 \times 10^{23} molecules 0.05 there are \frac{0.05}{18} \times 6.023 \times 10^{23} = 1.67 \times 10^{21} molecules$

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