Asked by Spenser
When 1.933 g of KNO3 is dissolved in 200 g of water initially at 23.57 ¢ªCelsius in a coffee-cup calorimeter, the final temperature is found to be 22.63 ¢ªCelsius. Calculate the heat released/absorbed per gram of KNO3 and per mol of KNO3 when KNO3 dissolves in water. NOTE: The heat capacity of calorimeter is 8.33 J ¢ªC-1. Assume the specific heat of the solution is the same as that of pure water
Answers
Answered by
DrBob222
KNO3(s) + H2O ==> KNO3(aq)
The temperature DECREASED so the solution process is endothermic. Heat is absorbed.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) + Ccal*(Tfinal-Tinitial)
q = [200g H2O x 4.184 J/g*C x (22.63-23.57)] + [8.33(22.63-23.57)]
q = -?. That's for 1.933 g; therefore,
-?/1.933 is q/gram KNO3 which is part a.
For part b and q/mol take q/gram x molar mass KNO3.
The temperature DECREASED so the solution process is endothermic. Heat is absorbed.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) + Ccal*(Tfinal-Tinitial)
q = [200g H2O x 4.184 J/g*C x (22.63-23.57)] + [8.33(22.63-23.57)]
q = -?. That's for 1.933 g; therefore,
-?/1.933 is q/gram KNO3 which is part a.
For part b and q/mol take q/gram x molar mass KNO3.
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