calculate the change in temperature when 47g of ammonium nitrate (NH4NO3) dissolve in 100g water. - need expla

Question

** I HAVE THE ANSWER, I NEED AN EXPLANATION BECAUSE I DON'T UNDERSTAND HOW TO DO THIS PROBLEM. THANK YOU!

calculate the change in temperature when 47g of ammonium nitrate (NH4NO3) dissolve in 100g water.

NH4+(aq) + NO3-(aq) --> NH4NH3(s) deltaH = -21.1 KJ/mol

MM(NH4NO3)=80.05g/mol
MM(H2O)=18.016g/mol
C(water)=75.3 mol^-1K^-1

a. 30K
b. 0.03K
c. -134K
d. -1K
e. -30K

Answer 1

OK...

You've been given and equation that tells you delta H for the formation of ammonium nitrate solid from its aqueous ions, so dissolving solid ammonium nitrate will absorb the same amount of heat per mole.

So, NH4NO3(s) --> NH4+(aq) + NO3-(aq) delta H = +21.1 kJ/mol.

You need to determine the number of moles of NH4NO3 you are dissolving. Divide the mass by it molar mass and you'll get 0.587 moles.

So, dissolving that much ammonium nitrate will absorb 0.587 mol X 21.1 kJ/mol = 12.4 kJ heat absorbed, or 12,400 Joules.

Now, you know that the heat capacity of water is 75.3 J/mol K. In other words, 75.3 Joules of heat will change the temperature of 1 mole of water by 1 K.

Calculate how many moles of water you have:

100 g / 18 g/mol = 5.56 mol water. So, to change the temperature of your 100 gram of water by 1 K, you will need:

5.56 X 75.3 = 419 J

Finally, if you divide the heat absorbed by 419, you'll find the change in temperature:

12,400 J/ 419 J/K= 29.6 K or 30 K. Since the dissolving absorbs heat, the water cools down, so your final answer is:
-30 K