Question

Calculate the standard enthalpy change for the dissolution of 10 grams of potassium hydroxide (KOH) in 100 mL of water. Given that the molar enthalpy of dissolution of solid KOH is -57.61 kJ/mol and the density of water is 1 g/mL.

Answer 1

To calculate the standard enthalpy change for the dissolution of 10 grams of potassium hydroxide  KOH  in 100 mL of water, we first need to determine the number of moles of KOH being dissolved. The molar mass of KOH is:K = 39.10 g/molO = 16.00 g/molH = 1.01 g/molTotal = 39.10 + 16.00 + 1.01 = 56.11 g/molNow, we can calculate the number of moles of KOH:moles of KOH = mass of KOH / molar mass of KOHmoles of KOH = 10 g / 56.11 g/mol = 0.1782 molNext, we can use the molar enthalpy of dissolution to find the enthalpy change for the dissolution of 0.1782 mol of KOH:H = moles of KOH  molar enthalpy of dissolutionH = 0.1782 mol   -57.61 kJ/mol  = -10.27 kJTherefore, the standard enthalpy change for the dissolution of 10 grams of potassium hydroxide  KOH  in 100 mL of water is -10.27 kJ.