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.

To calculate the heat of solution of LiI, you need to use the equation:

q = mcΔT

Where:
q is the heat absorbed or released by the solution (in joules),
m is the mass of the water (in grams),
c is the specific heat of water (in J/g·°C), and
ΔT is the change in temperature (in °C).

First, let's calculate the mass of water using the given values:

Mass of water = 57.5 g

Next, let's calculate ΔT, which is the change in temperature:

ΔT = Final temperature - Initial temperature
ΔT = 37.13°C - 21.40°C
ΔT = 15.73°C

Now, we can calculate the heat absorbed by the water (q) using the equation:

q = mcΔT

But since we're given the specific heat in J/g·°C, we need to convert the mass of water to grams:

Mass of water = 57.5 g

q = (57.5 g) × (4.184 J/g·°C) × (15.73°C)
q = 3785.94 J

Since the question asks for the heat of solution in kJ/mol, we need to convert J to kJ:

q = 3785.94 J = 3.78594 kJ

Finally, we need to calculate the heat of solution in terms of the number of moles of LiI. The molar mass of LiI is 133.84 g/mol (6.941 g/mol for Li + 126.90 g/mol for I-).

Using the molar mass of LiI, we can calculate the number of moles of LiI in the 8.00 g sample:

Moles of LiI = Mass of LiI / Molar mass of LiI
Moles of LiI = 8.00 g / 133.84 g/mol
Moles of LiI = 0.0597 mol

Finally, we can calculate the heat of solution of LiI in kJ/mol:

Heat of solution = q / Moles of LiI
Heat of solution = 3.78594 kJ / 0.0597 mol

Therefore, the heat of solution of LiI is approximately 63.51 kJ/mol.