Consider the dissolution of CaCl₂

Question

CaCl₂(s) → Ca²⁺(aq) + 2Cl⁻(aq) ∆H = -81.5 kJ/mol A 15.0-g sample of CaCl₂ is dissolved in 105.0 g water with both substances at 25.0°C. Calculate the final temperature (in °C) of the solution, assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.184 J/°C･g.

Answer 1

The thermochemical equation for the dissolution of CaCl₂:

CaCl₂(s) → Ca²⁺(aq) + 2 Cl⁻(aq)  ∆H = -81.5 kJ/mol

We can calculate the heat released by the dissolution (Q_d) of 15.0 g of CaCl₂

1. 81.5 kJ are released by the dissolution of 1 mole of CaCl₂.
2. The molar mass of CaCl₂ is 110.98 g/mol.

Q_d = (15.0 g / 110.98 g/mol) x ( -81.5 kJ/mol) = -11.0 kJ

According to the law of conservation of energy, the sum of the heat released by the dissolution (Q_d ) and the heat absorbed by the solution (Q_s) is zero.

Q_d + Q_s = 0

Q_s = - Q_d = 11.0 kJ

We can calculate the final temperature of the solution using the following expression

Q_s = m x c x (T_f - T_i)

T_f = (Qs/ mc) + T_i = [11.0*10^3 J / [ (105.0 g + 15.0 g) x 4.184 J/°C･g]] + 25.0°C = 46.9°C

m : mass of the solution

Answer: The final temperature of the solution is 46.9°C