Calculate the enthalpy of solution for the dissolution of lithium iodide, LiI, molar mass = 133.85 g mol^-1. When 1.49 g of LiI is dissolved in a coffee cup calorimeter containing 75.0 mL of water the temperature increased from 23.5 °C to 25.7 °C. The specific heat of water is 4.184 J g^-1 °C^-1, and assume the density of the solution is 1.00 g mL^-1.

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat absorbed by the solution:

q = m*s*ΔT

where,
m = mass of solution = 1.49 g + 75.0 g = 76.49 g
s = specific heat of water = 4.184 J/g°C
ΔT = change in temperature = 25.7°C - 23.5°C = 2.2°C

q = 76.49 g * 4.184 J/g°C * 2.2°C = 717.3 J

Next, we need to convert this amount of heat to the enthalpy of solution:

ΔH = q/n

where,
n = number of moles of solute = mass/molar mass = 1.49 g/133.85 g/mol = 0.01115 mol

ΔH = 717.3 J / 0.01115 mol = 64399.2 J/mol

Since enthalpy is usually expressed in kJ/mol, we convert J to kJ:
ΔH = 64.399 kJ/mol

Finally, since the enthalpy change for dissolving an ionic compound in water is usually negative (exothermic), we must multiply our answer by -1 to get the correct sign:

ΔH = -64.399 kJ/mol

The closest answer choice is C, -62 kJ/mol, which is the best choice for this problem.

Calculate the Enthalpy of Solution for the Dissolution of Lithium