(adsbygoogle = window.adsbygoogle || []).push({}); 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 H_{2}O) 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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Molar Heat of Solution

**Physics Forums | Science Articles, Homework Help, Discussion**