Question
9.0 g of KCl was dissolved in 300 mL of water in a calorimeter. Calculate the heat of solution in kJ/mol if the temperature dropped by 2.53 K on dissolution. The specific heat of water is 4.184 J/g K
Answers
DrBob222
I suppose you assume the density of H2O is 1 g/mL, then 300 mL has a mass of 300 g.
q = [mass H2O x specific heat H2O x delta T]
That q, in Joules, is caused by 9.0 g KCl
So heat solution is + (it gets colder) is q/9.0 g. Change that to kJ/g and that x molar mass KCl changes to kJ/mol.
q = [mass H2O x specific heat H2O x delta T]
That q, in Joules, is caused by 9.0 g KCl
So heat solution is + (it gets colder) is q/9.0 g. Change that to kJ/g and that x molar mass KCl changes to kJ/mol.