# Calculating enthalpy of dissolution

In a polystyrene cup calorimeter, $4.3\ \mathrm g$ of ammonium nitrate, $\ce{NH4NO}$, was added to $60.0\ \mathrm g$ of water and stirred to dissolve the solid completely. The initial temperature dropped from $22.0\ \mathrm{^\circ C}$ to a final temperature of $16.9\ \mathrm{^\circ C}$.

Calculate the enthalpy change in $\mathrm{kJ\ mol^{-1}}$ for this dissolution process, as represented by the chemical equation below:

$$\ce{NH4NO3(s) -> NH4NO3(aq)}$$

Assume that the calorimeter does not absorb any heat, that the density of the solution is the same as that of water $(1\ \mathrm{g\ ml^{-1}})$ and that the specific heat capacity of the solution is also the same as that of water $(4.18\ \mathrm{J\ g^{-1}\ K^{-1}})$.

$$M(\ce{NH4NO3})=80.05\ \mathrm{g\ mol^{-1}}$$

My attempt:

$$Q=mcT = 64.3\ \mathrm g \times 4.18\ \mathrm{J\ g^{-1}\ K^{-1}} \times -5.1\ \mathrm K = -1.37\ \mathrm{kJ}$$

But this doesn't seem right because I haven't used all the information given in the question, such as the molar mass of ammonium nitrate.

• So you have just worked out how much energy was taken in by the system in the dissolution process, 1.37 kJ. That's for 4.3g of ammonium nitrate. You are asked for kJ mol-1, so you need to find out how many moles of ammonium nitrate , that's why you have the molar mass of ammonium nitrate – Leeser Oct 28 '15 at 20:53
• Hi, just some notes on good practice: 1) You should write $\Delta T$ instead of $T$; the $\Delta$ sign represents a change in a quantity, in this case, temperature. 2) I think it is usually a good idea to put parentheses around your quantities, especially the negative one. Anyway, regarding the answer itself, @Leeser has already given you a nice hint, so see if you can work it out. – orthocresol Oct 29 '15 at 16:34

The question requires you to put your answer in kJ $mol^{-1}$. This means kilo joules per mole. So you have calculated the energy in kJ correctly, to get it per mole you need to first calculate the number of moles of ammonium nitrate you used.

Use the equation: $$n=\frac{m}{M_r}$$

n = moles

m = mass in grams

$M_r$ = molar mass

The mole calculation turn out to be 0.05371642723.

Then its just a matter of division, which I calculated to be -25.50 kJ $mol^{-1}$ (2.d.p).

That answer is using your rounded answer for the kJ, so the answer may be off. When i do calculations, i round right at the end to get the most accurate answer.

• not quite correct, you need to get the answer in kJ mol-1, hence divide by number of moles of ammonium nitrate in 4.3g – Leeser Oct 28 '15 at 20:56
• ah yes i missed the mol-1. what an idiot. sorry OP – Tahminul Oct 28 '15 at 20:57
• As you have realised, this answer as it stands is incorrect. I suggest you edit the extra information into your answer, but not before giving OP some time to try and see whether they can work it out themselves. – orthocresol Oct 29 '15 at 16:39