(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

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# Molar Heat of Solution

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