Calculate the enthalpy change (ΔH) for the dissolution of 10 grams of NaOH in 100 mL of water, given that the final temperature after mixing is 28°C and the initial temperature of water was 25°C. The specific heat capacity of water is 4.18 J/g°C. The molar mass of NaOH is 40.00 g/mol and the enthalpy of dissolution for NaOH is -44.51 kJ/mol.

Question

Calculate the enthalpy change (ΔH) for the dissolution of 10 grams of NaOH in 100 mL of water, given that the final temperature after mixing is 28°C and the initial temperature of water was 25°C. The specific heat capacity of water is 4.18 J/g°C. The molar mass of NaOH is 40.00 g/mol and the enthalpy of dissolution for NaOH is -

Answer 1

First, we need to calculate the number of moles of NaOH dissolved:moles of NaOH = mass of NaOH / molar mass of NaOHmoles of NaOH = 10 g / 40.00 g/mol = 0.25 molNow, we can calculate the enthalpy change for the dissolution of 0.25 mol of NaOH:H = moles of NaOH * enthalpy of dissolution per moleH = 0.25 mol *  -44.51 kJ/mol  = -11.1275 kJNext, we need to calculate the heat absorbed by the water:mass of water = 100 mL * 1 g/mL = 100 g  since the density of water is approximately 1 g/mL T = final temperature - initial temperatureT = 28C - 25C = 3Cq = mass of water * specific heat capacity of water * Tq = 100 g * 4.18 J/gC * 3C = 1254 J = 1.254 kJSince the dissolution of NaOH is exothermic, the heat absorbed by the water is equal to the heat released by the dissolution of NaOH:H = -q-11.1275 kJ = -1.254 kJHowever, this is not possible, as the enthalpy change for the dissolution of NaOH should be more negative than the heat absorbed by the water. There might be an error in the given data or the final temperature after mixing. Please check the data and try again.