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:
$$\ce{NH4NO3(s) -&gt; NH4NO3(aq)}$$
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.
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.