Question

Compute for the enthalpy (heat) of dissolution of a salt (NH₄Cl) in Joules using the following data:

• Mass of salt, g = 5.00
• Mass of calorimeter, g = 5.10
• Mass of calorimeter + water, g = 30.204
• Initial Temperature, °C = 25.35
• Final Temperature. °C = 17.19

Use the following:

• MW_NH4Cl = 53.5 g/mol
• Specific Heat_NH4Cl = 1.57 J/g·°C

Answer 1

Formula is Q = mass of water . specific heat of water . change in Temperature to calculate amount of heat gained by water.

• Mass of calorimeter, g = 5.10
• Mass of calorimeter + water, g = 30.204
• Mass of water = (30.204 - 5.1)g = 25.104 g

Q = 25.104 g . 4.18 J / g^oC. (17.19 - 25.35)^oC = -856.27 J

Water has lost -856.27J of heat energy to the salt, or salt has gained + 856.27 J of heat energy from water. So energy gained by salt is + 856.27 J.

Mass of salt, g = 5.00 and MW_NH4Cl = 53.5 g/mol

Moles of salt = (5 g / 53.5 g/mol) = 0.093 mol

Now, Change in enthalpy = Q gained by salt / # of moles of salt

= 856.27 J / 0.093 mol = 9162.089 J/mol = 9.162 kJ/mol.