How would you solve the enthalpy of reaction for 1.0 mole solution AB(aq) using this data?

Question

A(aq)+B(aq)-->AB(aq)
Info:
Specific heat=4.18J/(g*'C)
Density= 1.02g/mL
Solution A, =25.0 mL, 0.60 M, 21.4'C
Solution B, =25.0 mL, 0.60 M, 21.4'C
Solution AB, =1.0 mole, 25.3'C

Answer 1

Let us consider 1L of each of solutions A and B (it really does not matter if the volume is 25ml or not). You can perform the calculation with 25ml of A and B to varify my result.

By mixing 1L of A and 1L of B, we get 2L of mixture with 0.30M of A and 0.30M of B to react, forming 0.30M AB, a total 0.6 mole of AB.
The total heat released is:
(2000ml)*(1.02g/mL)*(4.18J/g*'C)*(25.3-21.4) = 33.3kJ
33.3kJ/0.6mol = 55.4 kJ/mol (answer)