Question

ΔH_solution of NaOH Phase Change:

NaOH(s)+H2O(l)-->NaOH(l)

Mass of NaOH: 1.98 g
ΔT for reaction A : 5.5 °C
Volume of water: 50ml

1.Calculate the value (calories) for the heat of solution.

2.Calculate the ΔH_solution for sodium hydroxide.

Answer 1

Solution :-

Volume of water = 50 ml

Density of water is 1 g/ml

Delta T for reaction = 5.5 C

Mass of water = volume x density

                            = 50 ml * 1 g/ml

                            = 50 g

Mass of solution = mass of water + mass of NaOH

                             = 50 g + 1.98 g

                              = 51.98 g

Part 1) Lets calculate the heat of solution.

q=m*c*delta T

= 51.98 g * 4.184 J per g C * 5.5 C

=1196 J

Converting joules to calorie

1196 J* 1 cal / 4.184 J = 286 cal

Heat is given off by the reaction therefore it has negative sign

Heat of solution = -286 cal

Part 2) calculating the delta H solution for the NaOH

Molar mass of NaOH = 40.0 g/mol

Delta H solution of NaOH = -286 cal * 40.0 g per mol / 1.98 g

                                               = -5777 cal/mol