Question Detail
Question
The ΔH for the solution process when solid sodium hydroxide dissolves in water is -44.4 kJ/mol. When a 13.9-g sample of NaOH dissolves in 250.0 g of water in a coffee-cup calorimeter, the temperature increases from 23.0°C to _____ °C. Assume that the solution has the same specific heat as liquid water, i.e., 4.18 J/g-K.
Correct Answer
Option D is the correct answer.
Detailed Explanation
Moles NaOH = 13.9 g / 40.0 g/mol = 0.3475 mol. Heat released = moles × ΔH = 0.3475 mol × 44.4 kJ/mol = 15.43 kJ = 15430 J. Total mass of solution = 13.9 + 250.0 = 263.9 g. q = mcΔT, so ΔT = q/(mc) = 15430 J / (263.9 g × 4.18 J/g°C) = 15430 / 1103 = 14.0°C. Final T = 23.0 + 14.0 = 37.0°C.
Hint
Calculate moles of NaOH, heat released, then use q = mcΔT.