i'm a bit confused in this problem. i think i know how to solve it, but i don't know if what i am doing is correct. please help.

here's the problem:

A 1.50g sample of NH

_{4}NO

_{3 (s)} is added to 35.0g of h

_{2}O

in a coffee cup calorimeter and stirred until it dissolves. The

temperature of the solution drops from 22.7

^{°}C to 19.4

^{°}C. What

is the heat of solution of NH

_{4}NO

_{3}, that is, what is the ΔH for

the process?

I tried using the formula q

_{rxn} = m*s*ΔT.

my value for m is the sum of the grams of ammonium nitrate and water, which is 36.5g.

my value for s is the sum of the specific heats of ammonium nitrate (1.77J/g°C) and water (4.186 J/g°C) which would be 5.956J/g°C.

ΔT would be the difference of 19.4°C and 22.7°C.

therefore, the value for q would be 220.694J.

since ΔH = q

_{rxn} / n

_{solute} ,

ΔH = 220.694J / (1.50g / 80.04g/mol)

= 11.8 kJ/mol

am i correct? thanks...