Ammonium Nitrate has a molar heat of dissolution of +25.8 kJ. What would the final temperature be of a solution made by adding 17.3 g of solid ammonium nitrate to a 250 g of water where both were initially at 25.0 degree Celcius. (assume that the solution has a specific heat of 4.18 J/g degree Celcius)

I tried this formula
qAmmoniumNitrate = –[(m)(Cp)(∆T)] by setting it equal to 25.8 kJ.
to solve for final temperature and it isn't working out... :/

The possible answers are
24.6
20.0
19.7
1.9

Question

Ammonium Nitrate has a molar heat of dissolution of +25.8 kJ. What would the final temperature be of a solution made by adding 17.3 g of solid ammonium nitrate to a 250 g of water where both were initially at 25.0 degree Celcius. (assume that the solution has a specific heat of 4.18 J/g degree Celcius) 

 I tried this formula 
qAmmoniumNitrate = –[(m)(Cp)(∆T)]  by setting it equal to 25.8 kJ.
to solve for final temperature and it isn't working out... :/

The possible answers are 
24.6 
20.0
19.7
1.9

gebooors · Answer

First you need to calculate amount of moles 
Of ammonium nitrate. One mole Needs 25.8 kJ
From surrounding to dissolve in water. So it cools water.