The amount of 2.00 mol of a solid ionic compound was dissolved in 225 mL of water in an insulated container. The initial temp of water was 20 deg. C and the final temp. was 47.8 deg C. What is the delta heat Dissolution? Assume the heat absorbed by the container is negligible.

Question

The amount of 2.00 mol of a solid ionic compound was dissolved in 225 mL of water in an insulated container.  The initial temp of water was 20 deg. C and the final temp. was 47.8 deg C. What is the delta heat Dissolution?  Assume the heat absorbed by the container is negligible.

Redox · Accepted Answer

d=m/v v=225mL=225cm³ d(water)=1g/cm³ m=225g Q=mcΔT Q=225 × 4200×(47.8 –20) Q=225 × 4.2 × 27.8 Q=–26271j/mol Q=–26.271kj/mol 2mole ==> –26.271kj/mol 1mole==> xkj/mol x=–13.1355kj/mol

DrBob222 · Answer

heat = q = mass water x specific heat water x (Tfinal-Tinitial). That is the q/2 mol.

carla · Answer

This was a multiple choice question, with possible answers of -13.1 KJ/mol; -6.26 KJ/mol; -27.8 KJ/mol; or -26.2 KJ/mol.

I was confused as to how to set equation but now I'm confused because if I'm doing this right-none of these answers work.
Thanks!

Michael · Answer

For anyone in the future, this is because you have q=mcat is solving for Joules, so convert that to kJ and divide by 2 to get kJ/mol.