Question

Consider the dissolution of Calcium Chloride:

                                             CaCl2(s)---->Ca^+2(aq) + 2Cl−(aq)                                 ΔHrxn= -81.5kJ

An 12.0 g sample of calcium chloride is dissolved in 100.0 g of water, with both substance at 22.0 degree C.

Molar Mass: CaCl2= 110.98g.mol    Molar Mass H20= 18.02g/mol   Density of H20= 1.00g/mol Cs(soln)= 4.20 J/(g K)

Will the temperature of the solution go up or down? Explain.

Calculate q(reaction)

Answer 1

​​​​​​Answer :-

1. the temperature of the solution go up (increases)

2. q_reaction = - 8.810 kJ = -8810 J

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Explanation :-

The given dissolution of CaCl₂ is

CaC12(g) + Caleta + 2Cla,
....(1)   
AH m = -81.5kJ
  

Given :-

Csolution = 4.20(J/g.K)

MassOfCaCl2 = 129

Molar MassOfCaCl2 = 110.98g/mol

MassOf water = 100g

Mass Of Solution = Masso fwater + MassOfCaci,

ie.
Mass Of Solution = 100(g) + 12 9) = 112(9)

T₁​​​​​​ = 22 °C = (22 + 273) = 295 K​​​​​​

T₂ = ?

First calculate moles of CaCl₂

MassOfCaCl2 Moleso fCaCl2 = Molar MassOfCaCl2

ie.
12(9) Moleso fCaCl2 = 110.98(g/mol) = 0.1081(mol)

from reaction (1)

Heat of dissolution of 1 mol of CaCl₂ = -81.5 kJ

therefore,

Heat of dissolution of 0.1081 mol of CaCl₂ =

=   
-81.5(kJ) x 0.1081(mol) 2 = -8.810(kJ) 1 (mol)

ie. Heat of dissolution of 12 g CaCl₂ = -8.810 kJ ==8810 J

Now this amount of heat is absorbed by CaCl₂ solution

therefore,

solution = -Arrn

ie.
Isolution = 8810(J)

(since heat is absorbed it has positive sign)

Now,

Heat gained by solution is given by

solution) = Cg.m.AT

ie.
Isolution) = Ca.m. (T2-T)
(Here m = mass of solution)

by putting value we have

8810(I) = 4.20(J/g.K) x 112(9) T2 - 295(K)

​​​ie.   
8810(J) T2 - 295(K) = 420(J/g.K) x 1129)

​​​ie.  
T2 - 295(K) = 18.73(K)

i.e  
T2 = 295(K) + 18.73(K)

i.e
T2 = 313.73(K) = 40.73(OC)

Since, temperature of the solution is grater than initial temperature of CaCl₂ and water, the temperature of solution go up.