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 4.82 g of CsClO₄(s) are dissolved in 118.30 g of water, the temperature of the solution drops from 23.64 to 21.14 °C.
The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.82 J/°C.
Based on the student's observation, calculate the enthalpy of dissolution of CsClO₄(s) in kJ/mol.
Assume the specific heat of the solution is equal to the specific heat of water

Answer 1

We can use the following formula

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat (units J/g∙oC), ∆ is a symbol meaning "the change in"

∆T = change in temperature (oC Celcius)

heat absorbed by CsClO4 = heat released by water + Heat released by calorimeter

Let us the amount of heat released by water

Q = ? m = 4.82 g +118.30 g = 123.12 g c=   4.184 J/g∙oC ∆T = 23.64°C - 21.14 °C = 2.5  °C

Q = 123.12 g x 4.184 J/g∙oC x 2.5  °C

Q =  1,287.836 Joules

Heat released by calorimeter = 2.5  °C x 1.82 J/°C = 4.55 Joules

heat absorbed by CsClO4 = 1,287.836 J + 4.55 J =  1,292.38 J = 1.292 kJ

Molar mass of CsClO4 = 232.36 g/mol

mass of CsClO4 used = 4.82 g

Moles of CsClO4 used = 4.82 g / 232.36 g/mol = 0.02074 Moles

The enthalpy of dissolution of CsClO4(s) =  1.292 kJ / 0.02074 Mol = 62.29 kJ /mol