Molar Heat of Solution

Christina-

1. The problem statement, all variables and given/known data
When 5.00 g of NaOH_(s) are added to 100 g of water using a calorimeter (with Cp = 493.24 J/K), the temperature rises from 25.0 to 37.5 °C. Calculate the molar heat of solution.
[align=center]NaOH(s)--->Na_(aq)+OH_(aq)[/align]
Assume that the specific heat capacity of water is 4.18 J/gK and that of the NaOH_(aq) solution is the same.

2. Relevant equations
1) q_p = C_p∆T,
2) | Heat Lost by Hot Water | = | Heat Gained by Cold Water | + | Heat Gained By Calorimeter |

3. The attempt at a solution

Heat Gained By Solution:
(100 g H₂O) x (4.18 J/gK) x (37.5 - 25)K = 5486.25 J

Heat Gained By Calorimeter:
Using q_p = C_p∆T,
C_p∆T = (493.24 J/K) x (37.5 - 25)K = 6165.5 J


| Heat Lost | = 5486.25 J + 6165.5 J = 11 651.75 J = 11.65175 kJ

Molar mass of NaOH = (22.99 + 16.00 + 1.01) = 40
5.00 g NaOH x (1 mol ÷ 40g) = 0.125 mol

11.65175 kJ ÷ 0.125 mol = 93.214 kJ/mol



I was just wondering if I did that correctly. I don't actually know the correct answer, but one of my fellow students got a different answer and I'm unsure if I'm the one with the wrong answer or not.
If this isn't the way to find ∆H, could someone show me how or perhaps point me to a useful page on the internet?

Thanks

kmcg

you wrote;
(100 g H2O) x (4.18 J/gK) x (37.5 - 25)K = 5486.25 J
but this should actually be;
(100 g H2O) x (4.18 J/gK) x (37.5 - 25)K = 5225 J
or
(105 g H2O) x (4.18 J/gK) x (37.5 - 25)K = 5486.25 J

I don't know if this helps but it's what I could see immediately