# Question 7f28d

Jul 16, 2017

$2.99 \times {10}^{-} 23$ $\text{mL}$

#### Explanation:

We're asked to find the volume occupied by $1$ $\text{molecule}$ of $\text{H"_2"O}$.

We can find this through a few steps:

1. Convert molecules to moles via Avogadro's number

2. Convert moles to grams via molar mass

3. Convert grams to cubic centimeters (milliliters) via density

Step 1:

We'll first convert the one molecule of water to moles, using Avogadro's number:

1cancel("molecule H"_2"O")((1color(white)(l)"mol H"_2"O")/(6.022xx10^23cancel("molecules H"_2"O")))

= color(red)(1.66 xx 10^-24 color(red)("mol H"_2"O"

Step 2:

Now, using the molar mass of water ($18.02$ $\text{g/mol}$), we'll calculate the number of grams:

color(red)(1.66xx10^-24)cancel(color(red)("mol H"_2"O"))((18.02color(white)(l)"g H"_2"O")/(1cancel("mol H"_2"O")))

= color(green)(2.99xx10^-23 color(green)("g H"_2"O"

Step 3:

Finally, knowing that $1$ $\text{g}$ water occupies a volume of $1$ ${\text{cm}}^{3}$ (or $1$ $\text{mL}$, which is more conventional for liquids), we can calculate the volume occupied by one molecules of water:

color(green)(2.99xx10^-23)cancel(color(green)("g H"_2"O"))((1color(white)(l)"mL")/(1cancel("g H"_2"O")))

= color(blue)(2.99xx10^-23 color(blue)("mL H"_2"O"#