# What are (a) the number of molecules in a water drop whose mass is 0.05*g; (b) the number of hydrogen atoms in such a drop; and (c) the number of oxygen atoms?

We know that $18.01 \cdot g$ of water is a molar quantity that contains $6.022 \times {10}^{23}$ individual water MOLECULES......and thus we can answer each question quantitatively..........
$a .$ $\text{Molecules of water in a rain drop}$ $=$ $\frac{0.05 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 = 1.67 \times {10}^{21} \cdot \text{water molecules}$
$b .$ $\text{Hydrogen atoms in a rain drop}$ $=$ $\text{TWICE the number of water molecules}$ $2 \times 1.67 \times {10}^{21} \cdot \text{hydrogen atoms}$.
$c .$ $\text{Oxygen atoms in a rain drop}$ $=$ $\text{THE SAME NUMBER as the number of water molecules}$ $1.67 \times {10}^{21} \cdot \text{oxygen atoms}$.