# Question 05a9c

Jan 8, 2018

$0.67$ drops of water.

#### Explanation:

First, we need to find out how much the $4 \cdot {10}^{23}$ molecules of water weighs. To do this, we can use this equation:

$\text{number of molecules in sample ⋅ mass of one molecule}$
$\text{= mass of sample}$

The mass of one water molecule is $\text{18.02 amu}$, because:

1. We know water is ${H}_{2} O$.
2. By looking at the periodic table, we can find the masses of hydrogen and oxygen—$\text{1.008 amu}$ and $\text{16.00 amu}$, respectively.

(The unit used is $\text{amu}$ because the mass of one atom is too small to be dealt with in terms of grams.)
3. $\text{mass of 2 hydrogen atoms + mass of 1 oxygen atom}$
$\text{= mass of 1 water molecule}$
4. (2*"1.008 amu") + "16.00 amu" = "18.02 amu"

The equation at the beginning can now be applied:

$\text{number of molecules ⋅ mass of one molecule = mass of sample}$

$\left(4 \cdot {10}^{23}\right) \cdot \text{18.02 amu" = 7.208 * 10^24 "amu}$

As the mass of the drop of water was given in grams, $7.208 \cdot {10}^{24} \text{amu}$ also needs to be converted into grams.
$\text{1 amu}$ = $1.66 \cdot {10}^{-} 24 \text{g}$, so:
$7.208 \cdot {10}^{24} \text{amu" = "11.97g}$

Then, we just divide $\text{11.97g}$ by $\text{18g}$, the mass of the drop of water, to find how many drops can fit into $\text{11.97g}$.

("11.97g")/("18g")="0.67 drops"#