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
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