Question

Question #13d55

Answer 1

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`