A raindrop weighs 0.050 g. How many molecules of water are in one rain drop?

Apr 10, 2016

Approx. $2.8 \times {10}^{-} 3$ $\times$ ${N}_{A}$, where ${N}_{A} = \text{Avogadro's number}$, 6.022140857(74)×10^23 mol^(−1).

Explanation:

We work out (i) a molar quantity, and then (ii) multiply this molar quantity by ${N}_{A}$, $\text{Avogadro's Number}$.

By definition in $18.01$ $g$ of water, $1$ $m o l$ of water, there are $\text{Avogadro's Number}$ of MOLECULES; and again in terms of $\text{Avogadro's Number}$, ${N}_{A}$, there are ${N}_{A}$ $\text{O}$ atoms and $2 \times {N}_{A}$ $\text{H}$ atoms, there are also masses of $15.999 \cdot g$ $O$ and $2 \times 1.00794 \cdot g$ $H$. I use ${N}_{A}$ here as I would any other collective number, a dozen, or a gross, or bakers' dozen.

So in $0.050$ $g$ water, there are $\frac{0.050 \cdot g}{18.015 \cdot g \cdot m o {l}^{-} 1}$ $\times {N}_{A}$, $\approx$ $2.8 \times {10}^{-} 3$ $\times$ ${N}_{A}$ water molecules.

From where did I get the elemental molar masses? Did I and do you have to remember them?