Question

Question #0112a

Answer 1

`"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.

`color(blue)(|bar(ul(color(white)(a/a)"molality" = "moles of solute"/"kilogram of solvent"color(white)(a/a)|)))`

In your case, the target solution must have a molality of `"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

`"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: