Question

When 0.250 moles of LiCl are added to 200.0 g of water in a constant pressure calorimeter a temperature change of +11.08°C is observed. Given that the specific heat of the resulting solution is 4.184 J g-1 °C-1 and we can ignore the small amount of energy absorbed by the calorimeter, what is the molar enthalpy of solution (ΔHsol) for LiCl?

Answer 1

3.91 × 10⁴ J/mol

Step-by-step explanation:

We can calculate the heat of solution using the following expression.

Q = c × m × ΔT

where,

c is the specific heat capacity of the solution

m is the mass of the solution

ΔT is the change in the temperature

The mass of LiCl is:

`0.250mol.(42.39g)/(mol) =10.6g`

The mass of the solution is:

m = mLiCl + mH₂O = 10.6 g + 200.0 g = 210.6 g

Q = c × m × ΔT = (4.184 J g⁻¹ °C⁻¹) × 210.6 g × 11.08 °C = 9763 J

In a constant pressure calorimeter, the molar enthalpy of solution for LiCl is:

`\Delta H_(sol)=(9763J)/(0.250mol) =3.91 * 10^(4) J/mol`