[dondoko]

I have been working on the following problem and have not been able to come up to the right answer.
Problem: "In a coffee-up calorimeter, 1.50g of NH4NO3 is mixed with 77.0g of water at an initial temperature of 25.00C.  After dissolution of the salt, the final temperature of the calorimeter contents is 23.53C.  Assuming the solution has a heat capacity of 4.18J/Cg, and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of NH4NO3 in units of kJ/mol.

Using the equation q= ms(dT),
(1.50g+77.0g)(4.18J/Cg)(25C-23C)= 370.2J.

Due to the unit conversion necessary from J/g to kJ/mol,
370.2J/g x 80.05g/mol NH4NO3 x 1kJ/1000 = 29.6kJ/mol

This turned out to be wrong.  I have tried molar mass of NH4NO3 and H2O combined, but it also turned out to be wrong.

Until getting to this point, I actually have tried many other ways, all of which were wrong.  This very latest calculation seems to be the best one I came up so far, but it is apparently wrong, and I cannot figure out what I am missing here.

Your advise will be appreciated.

Thanks,

[Yggdrasil]

the final temperature of the calorimeter contents is 23.53C. 

[...]

(1.50g+77.0g)(4.18J/Cg)(25C-23C)= 370.2J.

[dondoko]

It was actually, typo...  But thanks for pointing out.

I let it sit for a while, and I have further considered the solution as follows:

Using the equation q= ms(dT),
(1.50g+77.0g)(4.18J/Cg)(25.0C-23.53C)= 482.35J.

1.50g HH4NO3 x 1mol NH4NO3/ 80.05g NH4NO3 = 0.01873 mol NH4NO3

Due to the unit conversion necessary from J/g to kJ/mol,
(482.35J/g) / 0.01873mol x (1kJ/1000J) = 25.75kJ/mol = 25.8 kJ/mol

The final value is positive because the water temperature decreased, which means that the heat is absorbed by the NH4NO3 (endothermic).

I think that this looks much better than the last one.  Am I heading to the right direction?

[enahs]

First:

ΔT = T_final - T_initial.

This will get your signs correct with convention (will not change values though). Always Final-Initial.

Your last approach looks correct.