Question #13d55

1 Answer
Nov 5, 2017

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 g cm"^(-3). If you use the fact that

"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