Question

How many grams of CaCl2 should be added to 50.0 g of water to make a solution in which the mole fraction of Cl1- is 0.150?

***I understand how this question should be set up, I am confused about a certain part of the set up. The correct set up is as follows:

0.150 = (2n mol Cl-)/ (3n mol ions + 2.78 mol of H2O) and solve for n;

I do not understand why the ratio within the problem for Cl- to total mol ions was set up as 2 mols Cl- : 3 mol ions and not 2 mol Cl- : 1 mol CaCl2....why total ions and not total CaCl2 molecules? Could someone please explain this to me?

Thanks!

Answer 1

Assume that we are dissolving x moles of CaCl₂ in water, we end up with 2x moles of Cl^- and x moles of Ca²⁺ ions in the solution. So, we have total moles of ions present in the solution are 2x Cl^- and x Ca²⁺, total 3x moles of ions present in the solution.We need to add no. of moles of water to get the total number of moles present in the solution (i.e; total number of moles = n_total = 2x Cl^- + x Ca²⁺ + n_H2O)

no. moles of Cl^- = 2x

mole fraction Cl^- = X_Cl^- = (no. moles of Cl^- )/(2x Cl^- + x Ca²⁺ + n_H2O)

Since, we have mole fraction of Cl^- (given), we can find out how many moles of CaCl2 needed to make the desired solution. When we deal with mole fractions, we count no. of different species present in the solution. When we dissolve CaCl2 in aqueous solution, it is not in the form of CaCl2 molecule but in the form of Ca2+ and Cl- ions. That's why we look for the number of ions.