posted by Sug on .
LiI(s) → Li+(aq) + I‾ (aq)
Solid LiI dissolves as shown in the above equation. A 8.00 g sample of LiI was dissolved in 57.5 g of water. The initial temperature of the water was 21.40°C. After the compound dissolved, the temperature of the water was 37.13°C. Assume the heat was completely absorbed from the water and no heat was absorbed by the reaction container or the surroundings. Calculate the heat of solution of LiI in kJ/mol. The specific heat of water is 4.184 J/g·°C. (Do not take the added mass of the LiI into account when calculating q for the solution.)
Can someone explain how to do this?
q = heat generated = mass water x specific heat water x delta T.
Convert to kJ.
Then kJ/8 = kJ/gram. Then convert that to kJ/mol. Post your work if you get stuck.