Question

What is the standard enthalpy change of dissolution for 5.00 grams of solid sodium hydroxide (NaOH) in 100.0 mL of water at 25°C? Given: The molar enthalpy of dissolution of NaOH is -44.51 kJ/mol and the density of water is 0.997 g/mL.

Answer 1

To calculate the standard enthalpy change of dissolution for 5.00 grams of solid sodium hydroxide  NaOH  in 100.0 mL of water at 25C, we need to follow these steps:1. Calculate the moles of NaOH.2. Use the molar enthalpy of dissolution to find the enthalpy change.Step 1: Calculate the moles of NaOHThe molar mass of NaOH = 22.99  Na  + 15.999  O  + 1.007  H  = 39.996 g/molmoles of NaOH = mass / molar massmoles of NaOH = 5.00 g / 39.996 g/mol = 0.125 molStep 2: Use the molar enthalpy of dissolution to find the enthalpy changeH = moles  molar enthalpy of dissolutionH = 0.125 mol   -44.51 kJ/mol H = -5.56 kJThe standard enthalpy change of dissolution for 5.00 grams of solid sodium hydroxide  NaOH  in 100.0 mL of water at 25C is -5.56 kJ.