# Question #21afc

##### 1 Answer

#### Explanation:

The idea here is that you need to use the **density** of water to convert that *volume* to *mass*.

Now, because the problem didn't provide you with a value for the density of water, or at least with the *temperature* at which the water is being kept, you can assume the density to be equal to

#rho_("H"_ 2"O") = "1 g cm"^(-3)#

In your case, the volume of water is expressed in *cubic decimeters*, *cubic centimeters*,

#1.00 color(red)(cancel(color(black)("dm"^3))) * (10^3color(white)(a)"cm"^3)/(1color(red)(cancel(color(black)("dm"^3)))) = 1.00 * 10^3"cm"^3#

Use the density of water to convert this to *grams*

#1.00 * 10^3 color(red)(cancel(color(black)("cm"^3))) * "1 g"/(1color(red)(cancel(color(black)("cm"^3)))) = 1.00 * 10^3"g"#

Your last step here will be to use the **molar mass** of water to convert the *grams* of water to **moles**

#1.00 * 10^3color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("55.5 moles H"_2"O")color(white)(a/a)|)))#

The answer is rounded to three **sig figs**, the number of sig figs you have for the volume of water.