When 4.21 grams of potassium hydroxide are added to 250 mL of water in a coffee cup calorimeter, the temperature rises by 4.14°C.

Assume that the density and specific heat of the dilute aqueous solution are the same as those of H2O and calculate the molar heat of solution of potassium hydroxide in kJ/mol. C = 4.184 J/g*K

The molar heat of solution of potassium hydroxide = 57.7 kJ / mol

Explanation:

Step 1: Data given

Mass of potassium hydroxide = 4.21 grams

Volume of water = 250 mL

Temperature rise = 4.14 °C

Density = 1g/mL

Specific heat = 4.184 J/g°C

Step 2: Calculate the heat absorbed by water

q = m*c*ΔT

⇒ with m = the mass of water = 250 grams

⇒ with c = the specific heat of the solution = 4.184 J/g°C

⇒ with ΔT = 4.14 °C

q = 250 * 4.184*4.14 = 4330.4 J

Step 3: Calculate moles of KOH

Moles KOH = Mass KOH / Molar mass KOH

Moles KOH = 4.21 grams / 56.106 g/mol

Moles KOH = 0.075 moles

Step 4: Calculate molar heat of solution

4330.4 J / 0.075 moles = 57738.7 j/mol = 57.7 kJ / mol

(Note that the enthalpy change for the reaction is negative because the reaction is exothermic)

The molar heat of solution of potassium hydroxide = 57.7 kJ / mol