To calculate the enthalpy change of the dissolution of 5 grams of sodium hydroxide NaOH in 100 mL of water, we first need to determine the number of moles of NaOH in 5 grams.The molar mass of NaOH is:Na = 22.99 g/molO = 16.00 g/molH = 1.01 g/molTotal = 22.99 + 16.00 + 1.01 = 40.00 g/molNow, we can find the number of moles of NaOH in 5 grams:moles = mass / molar massmoles = 5 g / 40.00 g/mol = 0.125 molNext, we can use the molar enthalpy of dissolution to find the enthalpy change for the dissolution of 0.125 mol of NaOH. The molar enthalpy of dissolution is -44 kJ/mol, which means that -44 kJ of energy is released per mole of NaOH dissolved.Enthalpy change = moles molar enthalpy of dissolutionEnthalpy change = 0.125 mol -44 kJ/mol = -5.5 kJTherefore, the enthalpy change of the dissolution of 5 grams of sodium hydroxide NaOH in 100 mL of water is -5.5 kJ.