Question

When 1.3 g of KClO3 was added to 151.1 g of water in a calorimeter, the  temperature dropped by 0.588 °C. The heat capacity of H2O is 4.184 J/g°C. Assume the  specific heat of the solution equals that of pure H2O and that the calorimeter neither  absorbs nor leaks heat. What is the molar heat of solution of solid potassium chlorate? Give the answer in kJ.

Answer 1

mass of KClO3 = 1.3 g

moles of KClO3 = 1.30 / 122.55 = 0.0106 mol

mass of solution = 151.1 + 1.3 = 152.4 g

temperature dropped = 0.588 oC

Cp = 4.184 J / g oC

Q = m Cp dT

= 152.4 x 4.184 x - 0.588

Q = -375 J

delta H = - Q / n

            = 0.375 / 0.0106

delta H = 35.37 kJ/mol

molar heat of solution = 35.4 kJ