5.00 grams of sodium hydroxide was placed in 100.0 grams of water in a coffee cup calorimeter. The temperature of the water increased from 25.0 deg C to 35.2 deg C. Find the heat of solution (ΔHsoln) for sodium hydroxide in kJ/mol. (Specific heat of water = 4.184J/g degC)?

The heat of solution, ΔHsoln, of  NaOH is -34.139 kJ/mol

Explanation:

Here we have

Mass of sodium hydroxide = 5.00 g

Mass of water = 100.0 g

Initial temperature of water, T₁ = 25 °C

Final temperature of water, T₂ = 35.2 °C

Temperature rise of water, ΔT= T₂ - T₁ = 35.2 - 25 = 10.2 °C

Therefore, from the equation of heat capacity of a medium, we have

Q = m·c·ΔT

Where;

Q = Heat acquired by water

m = Mass of water present = 100.0 g

ΔT = 10.2 °C

c = Specific heat capacity of water = 4.184 J/(g·°C)

Therefore, by plugging in the values we have;

Q = 100 g × 4.184 J/(g·°C) × 10.2 °C = -4267.68 J

Molar mass of NaOH = 39.997 g/mol

Therefore number of moles of NaOH present in 5.00 g = 5/39.997 = 0.125 moles

Which means that the heat of solution, ΔHsoln, of 0.125 moles of NaOH = -4267.68 J

Therefore, the heat of solution of 1 mole of NaOH =

= -34138.9 J/mole  =

= -34.139 kJ/mol

The heat of solution, ΔHsoln, of  NaOH = -34.139 kJ/mol.

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error