# Question 21afc

Jun 29, 2016

$\text{55.5 moles H"_2"O}$

#### 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, ${\text{dm}}^{3}$, so the first thing to do here is convert it to cubic centimeters, ${\text{cm}}^{3}$

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.