Assume the specific heat of the solution is equal to the specific heat of water.

Question

When a solid dissolves in water, heat may be evolved or absorbed. The heat of dissolution (dissolving) can be determined using a coffee cup calorimeter.

In the laboratory a general chemistry student finds that when 10.91 g of NaClO₄(s) are dissolved in 104.10 g of water, the temperature of the solution drops from 23.84 to 20.56 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.78 J/°C.

Based on the student's observation, calculate the enthalpy of dissolution of NaClO₄(s) in kJ/mol.

Assume the specific heat of the solution is equal to the specific heat of water.

Answer 1

You should use the following formula:

q = mC∆T + C_cal∆T

q= heat = ?

m = mass = 104.10 g (most intro courses do not include mass of solute, but you may if answer is wrong)

C = specific heat of water = 4.184 J/gº

∆T = change temperature = 20.56 - 23.84 = -3.28º

C_cal = calorimeter constant = 1.78 J/º

Solving for q ...

q = (104.10 g)(4.184 J/gº)(-3.28º) + (1.78 J/º)(-3.28º)

q = -1429 J + -5.839

q = -1434 J

Molar mass NaClO4 = 122.4 g / mol

moles NaClO4 present = 10.91 g x 1 mol / 122.4 g = 0.08913 moles

∆H = -1434 J / 0.08913 mols x 1 kJ / mol = -16.09 kJ/mole