Calculating Molar Heat of Solution for NaOH using a Calorimeter | Homework Help

[Christina-]

Homework Statement

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.

Homework Equations

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

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]

One small thing I can see immediately

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

 

[Blueshift5]

for your help!



Your calculations look correct to me. The molar heat of solution for NaOH using a calorimeter should be 93.214 kJ/mol. It's possible that your classmate made an error in their calculations. However, it's always good to double check your work and make sure you are using the correct equations and values. Additionally, it's always a good idea to compare your answer to a known value or do some research to see if your answer is within a reasonable range. Keep up the good work!