# Question c8ba4

Nov 25, 2016

$\text{56 moles H"_2"O}$

#### Explanation:

We don't have a direct connection between moles and volume, but we do have one between moles and mass.

This means that your first step will be to convert the volume of water to grams by using water's density.

Since you only have one significant figure for the volume, it would make sense to use the approximated value of ${\text{1 g mL}}^{- 1}$ for the density.

Since you know that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 L" = 10^3"mL}}}}$

you will have

1 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = 10^3"g"

Now, water has a molar mass of ${\text{18.015 g mol}}^{- 1}$, which means that every mole of water has a mass of $\text{18.015 g}$.

You can use the molar mass as a conversion factor to see how many moles would be present in your sample

10^3color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"#

I will leave the answer rounded to two sig figs, but keep in mind that you only have one sig figs for the volume of the sample

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{no. of moles H"_2"O} = 56}}}$