Question

Calculate the heat of crystallization of NaOH given that 10 grams of NaOH dissolves in 100 mL of water at 25°C and the resulting solution is cooled to 0°C. The enthalpy of solution of NaOH is -44.5 kJ/mol, and the density of the resulting solution is 1.1 g/mL.

Answer 1

To calculate the heat of crystallization of NaOH, we need to consider the heat absorbed or released during the dissolution process and the heat absorbed or released during the cooling process.First, let's calculate the moles of NaOH dissolved:Moles of NaOH = mass of NaOH / molar mass of NaOHMoles of NaOH = 10 g /  22.99 g/mol + 15.999 g/mol + 1.007 g/mol Moles of NaOH = 10 g / 39.996 g/molMoles of NaOH = 0.250 molNow, let's calculate the heat absorbed or released during the dissolution process:Heat of dissolution = moles of NaOH  enthalpy of solutionHeat of dissolution = 0.250 mol   -44.5 kJ/mol Heat of dissolution = -11.125 kJNext, let's calculate the heat absorbed or released during the cooling process. We will use the formula:q = mcTwhere q is the heat absorbed or released, m is the mass of the solution, c is the specific heat capacity of the solution, and T is the change in temperature.First, we need to find the mass of the solution:Mass of solution = volume of solution  density of solutionMass of solution = 100 mL  1.1 g/mLMass of solution = 110 gNow, let's assume the specific heat capacity of the solution is similar to that of water, which is 4.18 J/gC. We can now calculate the heat absorbed or released during the cooling process:q = mcTq =  110 g  4.18 J/gC  25C - 0C q =  110 g  4.18 J/gC  25C q = 11485 Jq = 11.485 kJSince the heat absorbed during the cooling process is positive, the heat released during the dissolution process must be equal to the heat absorbed during the cooling process for the solution to cool down to 0C.Heat of crystallization = - Heat of dissolutionHeat of crystallization = 11.125 kJTherefore, the heat of crystallization of NaOH is 11.125 kJ.