1. Consider these two solutions: Solution A is prepared by dissolving 5.00 g of

Question

1. Consider these two solutions: Solution A is prepared by dissolving 5.00 g of MgCl2 in enough water to make 0.250 L of solution, and Solution B is prepared by dissolving 5.00 g of KCl in enough water to make 0.250 L of solution. Which direction will solvent initially flow if these two solutions are separated by a semipermeable membrane?

2. Assuming that the volumes of the solutions described in question #1 are additive and ignoring any effects that gravity may have on the osmotic pressure of the solutions, what will be the final volume of solution A when the net solvent flow through the semipermeable membrane stops?

Explanation / Answer

Lets first define osmotic pressure. Osmotic pressure is the minimum pressure which needs to be applied to a solution to prevent the inward flow of water across a semipermeable membrane. It is also defined as the measure of the tendency of a solution to take in water by osmosis. The solvent will flow from the solution which is more dilute as they will have less osmotic pressure. now osmotic pressure is a colligative property, so it depends on the molality of the non volatile solute and not property of solute. in both the sotution A and B we have 250 mL of solution and same mass of solute, but 1 mol of MgCl2 dissocialtes into one mol of Mg2+ and two mols of Cl-, so there are 3 moles of solute, so has higher molality than the solution 2, since it has only 2 moles of ions (one mol of K+ and one mol of Cl-, for each mol of KCl).

now let's calculate the concentrations of two solutions.

Taking into account of the dissociated ion, in solution A initially mols of MgCl2 present = mass/molar mass = 5/95 = 0.0526 moles, so the no. of dissociated moles = 0.0526 X 3 = 0.158 moles

therefore molarity = 0.158/0.25 = 0.63 M

similarly moles of KCl in solution B = 5/74.5 = 0.067 moles,

so moles of ions = 0.067 X 2 = 0.134 moles

therefore molarity of solution B = 0.134/0.25 =0.537 M

So the solution A has higher concenration (0.63 M) than solution B (0.537M), and so solution A has higher osmotic pressure, and water will move from solution B to Solution A till the equilibrium is reached, i.e. both the containers have same concentration.

2. To find the final volume in solution A, we have to find the concenration at equilibrium.

concentration at equilibrium = (0.63 + 0.537)/2 = 0.58M

so in container A moles of dissociated solute ions = 0.158 moles,

so the final volume of solvent when the equilibrium i.e. concentration reaches 0.58 M = 0.158/V

or V = 0.158/0.58 = 0.272 L

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