A student weighed out 1.74 grams of sodium hydroxide pellets andadded them to 48

Question

A student weighed out 1.74 grams of sodium hydroxide pellets andadded them to 48.0 grams of water in a calorimeter. Extrapolatingback to the time of mixing yielded a temperature change at the timeof mixing equal to 9.20 degrees C. Calculate the heat of solutionper mole of NaOH in J. Assume that the specific heat of the sodiumhydroxide solution is 3.93 J/K.g. Enter your answer as a numberwith three significant figures.



This is what I did...but it was wrong....
q=(total mass)(specific heat)(T)
q=(49.74g)(3.93J/K*g)(282.2K)=55163.94804J

mol of solute=1.74g*(mol/40g)=0.0435mol

Hsoln=q/mol solute
Hsoln=55163.94804J/0.0435mol=1.27*10^6 J/mol

I tried both positive and negative signs (neither are right forthis answer).

Explanation / Answer

Given data Mass of solution = 48.0 g +1.74 g                            = 49.74 g Specific heat capacity, s = 3.93 J/K.g                                    T= 9.2 0C                                         = 9.2 K Heat of solution, q = ms T                               = (49.74g)(3.93 J/K.g)(9.2K)                               = 1798 J            The  heatof solution per mole of NaOH in J = 40 g/mol *1798J/1.71g                                                                              = 41342.5 J/mol

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