Question #0112a
1 Answer
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)|)))∣∣ ∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aamolality=moles of solutekilogram of solventaa∣∣∣−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
In your case, the target solution must have a molality of
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"_318.7g⋅1 mole NH4NO380.043g=0.2336 moles NH4NO3
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"0.2336moles NH4NO3⋅given molality1 kg water0.542moles NH4NO3=0.431 kg water
To convert this to grams, use the conversion factor
"1 kg" = 10^3"g"1 kg=103g
You will have
"mass of water" = color(green)(|bar(ul(color(white)(a/a)"431 g"color(white)(a/a)|)))mass of water=∣∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aa431 gaa∣∣−−−−−−−−−
The answer is rounded to three sig figs.
A cool video on molaity:
