When a 4.00 g sample of KSCN is dissolved in water in a calorimeter W/ total heat capacity of 2.85 kJ⋅K−1, the temperature decreases by 0.350 K. Calculate the molar heat of solution of KSCN.

Answer 1

q = Ccal∆T = 2.85 kJ/º * 0.35º = +0.9975 kJ (+ because the rx is endothermic)

moles KSCN = 4.00 g x 1 mol/97.18 g = 0.0412 moles

Molar heat of solution = +0.9975 kJ/0.0412 moles = 24.2 kJ/mol