Question

A 0.100-L solution is made by dissolving 0.441 g of CaCl2(s) in water. (a) Calculate the osmotic pressure of this solution at 27 °C, assuming that it is completely dissociated into its component ions. (b) The measured osmotic pressure of this solution is 2.56 atm at 27 °C. Explain why it is less than the value calculated in (a), and calculate the van’t Hoff factor, i, for the solute in this solution. (c) The enthalpy of solution for CaCl2 is ΔH = –81.3 kJ/mol. If the final temperature of the solution is 27 °C, what was its initial temperature? (Assume that the density of the solution is 1.00 g/mL, that its specific heat is 4.18 J/g-K, and that the solution loses no heat to its surroundings.)

Answer 1

a) The Osmotic pressure of the solution is given by the equation, Pie = MRT. (M = molarity). We know temperature, T = 270 C = 300 K, R= 8.314 K.Pa dm3 mol-1 K-1.We can calculate the molarity of the solution from the mass of CaCl₂ and volume pf the solution:

Molarity= (0.441g/111.0 g mol-1) * (1/0.100) = 0.0397 mol/dm³

When completely disassociated each CaCl₂ gives 3 ions, hence total concentration of ions in the solution = 3*0.0397 = 0.119 M.

The osmotic presure = MRT =(0.119 mol/dm³) (8.314 K.Pa dm3 mol-1 K-1) (300 K ) = 2.97*10² k.Pa

b)

here there is influenceof temperature change which will result in the difference between the calculated and measured.

Van't Hoff factor, i is the number of ions or particles formed from one formula unit of solute in solution,

Here one Ca ion and two Cl ions are formed ffrom one formula unit so here Van't Hoff Factor = 3

C) If the solution is 0.0397 M CaCl₂ and the has a total volume of 0.100 dm³ , the number pf moles of solute is = 0.00397 moles. Hencethe quantity of heat generated is -81.3KJmol^-1 * 0.00397 = -0.323 KJ. The solution absorbs the heat, causing the temperature to increase. The relationship between heat and temperatur change is,

q = (Specific heat) (grams) (temp Deifference)

Heat absorbe by the solution = 0.323 KJ = 323 J , Specific heat = 4.18 J/g-K, Grams = 100

Temperature difference = (323 J)/(4.18 J/g-K *100) = 0.773 K

Initial Temp was= 300 K - 0.773K = 299.2 K = 26.2 ⁰C