A 3.53g sample of (NH4NO3) was added to 80.0mlof water in a constant pressure calorimeter of negligible heat capacity. As a result the temperature of the water decreased from 21.6C to 18.1C Calculate the heat of solution of NH4NO3. Help me ....

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anonymous · Answer

The temperature change was 3.5 degrees C. The specific heat of water is 4.18 J/g C. To raise the temperature of 80ml of water by 3.5 degrees you need 3.5 * 80 * 4.18 = 1170 joules. The molecular weight of ammonium nitrate is 80g/mol, so you have 0.044125 mols. That means the heat of solution is 1170/0.04412 = 26518 = 26.5 kilojoules per mole.

anonymous · Answer

Thank you !! :)