[adenine135]

A student determined the enthalpy of NaOH by dissolving 4.00 g of NaOH(s) in 180 mL of pure water (density 0.999 g/mL) at an initial temperature of 19.5 °C. Temperature-time data collected after mixing was extrapolated back to the time of mixing to obtain a temperature change, ΔT = +5.5 °C. The density of the final solution was 1.013 g/ml at 25 °C, and the solution heat capacity was 4.08 J/(g °C). In a separate experiment, the calorimeter constant was found to be 21 J/°C.


What was the molar enthalpy of dissolution of NaOH in water?


So this is my work (I believe I solved the problem, just want verification):

mass of NaOH = 4.00 g
mass of water = 180 mL (0.999 g/ mL) = 179.82 g
T_i = 19.5 °C, T_f = 25 °C , ΔT = +5.5 °C
C_s = 4.08 J / (g °C)

4.00 g NaOH (1 mol NaOH / 39.9971 g NaOH) = 0.1000 mol NaOH
179.82 g H₂O (1 mol H₂O / 18.0153 g H₂O) = 9.9815 mol H₂O

ΔH = m Cs ΔT
     = (179.82 g + 4.00 g) [4.08 J / (g °C)] (+5.5 °C)
     = 4124.9208 J


So then molar enthalpy would be the enthalpy of the reaction divided by the amount of moles:

molar enthalpy = 4124.9208 J  / (0.1000 mol NaOH + 9.9815 mol)
                     = 410 J / mol

[Would this be the right amount of sig. figures since the temperature has only 2?]