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 2.21 g of CaBr₂(s) are dissolved in 108.50 g of water, the temperature of the solution increases from 25.16 to 27.72 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.70 J/°C.

Based on the student's observation, calculate the enthalpy of dissolution of CaBr₂(s) in kJ/mol.

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

Answer 1

ost by the water = m*c*dt

= mass*specific heat * temp change

(4.18 J/g·°C) x (108.50 g) x (27.72-25.16)°C = 1161.04 J lost by the water

Supposing the calorimeter to be at the same temperatures as the water:

(1.70 J/°C) x (27.72-25.16)°C = 4.352 J lost by the calorimeter

1161.04 J + 4.352 J = 1165.392 J total lost by the combined apparatus, and gained by the solute

Number of mole s= amount in g/ molar mass

= 2.21 g of CaBr₂(s)/ 199.89 g/mol

=0.011 mole CaBr2

enthalpy of dissolution of CaBr₂(s) in kJ/mol= Total heat / number of mole

= 1165.392 J /0.011 mole CaBr2

= 105944.7273 J/ mole *1 KJ/1000 J

= 105.94 KJ/ mole