# Question 0112a

Mar 19, 2016

$\text{431 g}$

#### Explanation:

First, make sure that you understand what it is you're looking for here.

A solution's molality will tell you how many moles of solute you get per kilogram of solvent.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{molality" = "moles of solute"/"kilogram of solvent} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, the target solution must have a molality of $\text{0.542 m}$. This si equivalent to saying that this solution contains "0.542 moles of ammonium nitrate, your solute, for every one kilogram of water.

In order to find the number of moles of ammonium nitrate, use the compound's molar mass

18.7 color(red)(cancel(color(black)("g"))) * ("1 mole NH"_4"NO"_3)/(80.043color(red)(cancel(color(black)("g")))) = "0.2336 moles NH"_4"NO"_3

Now, you can sue the molality of the solution as a conversion factor to help you determine how many kilograms of water would be needed in order to have the target molality

0.2336color(red)(cancel(color(black)("moles NH"_4"NO"_3))) * overbrace("1 kg water"/(0.542color(red)(cancel(color(black)("moles NH"_4"NO"_3)))))^(color(purple)("given molality")) = "0.431 kg water"

To convert this to grams, use the conversion factor

$\text{1 kg" = 10^3"g}$

You will have

"mass of water" = color(green)(|bar(ul(color(white)(a/a)"431 g"color(white)(a/a)|)))#

The answer is rounded to three sig figs.

A cool video on molaity: