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## 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 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