# Question #13d55

##### 1 Answer

Here's what I got.

#### Explanation:

Start by calculating the *mass* of water present in your sample. To do that, you need to use the **density** given to you.

You know that at a certain temperature, water's density can be approximated to

#"1 L" = "1 dm"^3" "# and#" " "1 dm"^3 = 10^3color(white)(.)"cm"^3#

you can say that your sample will have a mass of

#1 color(red)(cancel(color(black)("L"))) * (1 color(red)(cancel(color(black)("dm"^3))))/(1color(red)(cancel(color(black)("L")))) * (10^3 color(red)(cancel(color(black)("cm"^3))))/(1color(red)(cancel(color(black)("dm"^3)))) * overbrace("1 g"/(1color(red)(cancel(color(black)("cm"^3)))))^(color(blue)("the density of water")) = 10^3color(white)(.)"g"#

At this point, all you have to do is to use the **molar mass** of water to find the number of *moles* present in the sample.

#10^3 color(red)(cancel(color(black)("g"))) * overbrace(("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))))^(color(blue)("the molar mass of water")) = "55.51 moles H"_2"O"#

I'll leave the answer rounded to four **sig figs**, but keep in mind that you have one significant figure for the volume of water. So the answer *should* be reported as

#"no. of moles" = 56#