Answer:
26.85kJ/mol is the heat of dissolution of Cs₂O₄
Step-by-step explanation:
The process of dissolution of Cs₂SO₄ in water occurs as follows:
Cs₂O₄(s) → 2Cs⁺(aq) + SO₄²⁻(aq) + ΔH
Where ΔH is the change in heat per mole of Cs₂O₄
The first we can see is ΔH > 0 because heat is absorbed (Temperature is decreasing) when the reaction occurs.
Now, the change in heat of reaction is:
q = Heat calorimeter + Heat solution
q = ΔT*1.65J/°C + S*m×ΔT
Where q is the heat of reaction.
ΔT is change in temperature: 25.62°C - 22.57°C = 3.05°C
S is specific heat of solution = Specific heat water = 4.184J/g°C
m is mass of solution = 22.24g + 106.70g = 128.94g
Replacing, heat of reaction is:
q = 3.05°C*1.65J/°C + 4.184J/g°C*128.94g×3.05°C
q = 1650J are absorbed when 22.24g of Cs₂O₄ reacts
Moles of 22.24g of Cs₂O₄ are - Molar mass: 361.87g/mol-:
22.24g * (1mol / 361.87g) = 0.06146 moles
That means, when 0.06146 moles of Cs₂O₄ react, the heat absorbed is 1650J. That means the heat absorbed per mole of Cs₂O₄ (Enthalpy of dissolution) is:
1650J / 0.06146 moles = 26847J / mol =
26.85kJ/mol is the heat of dissolution of Cs₂O₄