# Question #f28a7

Dec 9, 2014

The staggering number of water drops is roughly $6.0 \cdot {10}^{23}$.

First, we need to find out the distance between Earth and the moon at perigee, which is the closest point to Earth the moon reaches in its orbit. This distance is listed at $225 , 804$ miles, which, after converting to $k m$, equals $363 , 396 k m$.

We can now determine the volume for our water column. Starting from its rectangular base of $9.2$ X $9.2 k m$, we get

${V}_{w a t e r} = 9.2 \cdot 9.2 \cdot 363 , 396 = 30 , 757 , 837 k {m}^{3}$

Each $k {m}^{3}$ has $1 \cdot {10}^{15} c {m}^{3}$, so the volume can be said to be

${V}_{w a t e r} = 30 , 757 , 837 \cdot {10}^{15} c {m}^{3}$.

Now, there are approximately $20$ drops in $1 c {m}^{3}$ of water, so the number of drops in a column this size should be

$n u m b e {r}_{\mathrm{dr} o p s} = \frac{20 \mathrm{dr} o p s}{1 c {m}^{3}} \cdot 30 , 757 , 837 \cdot {10}^{15} c {m}^{3} \to$

$n u m b e {r}_{\mathrm{dr} o p s} = 615 , 156 , 740 \cdot {10}^{15}$, or roughly $6.0 \cdot {10}^{23}$

I've just noticed how close to Avogadro's number, which is $6.022 \cdot {10}^{23}$ molecules per $1$ mole of substance...very good problem for trying to visualize and understand the concept of mole.