However, neither individual value is required.Ģa) We know the following: MgCl 2 yields 3 particles per formula unit and its molar mass is 95.211 g mol¯ 1. Using the provided data, we cannot individually get to either the van't Hoff factor or the analytical concentration. What is the mass percent of MgCl 2 in the solid? (Assume ideal behavior for the solution.)ġ) Calculate the apparent molarity of the solution:Ĭomment: in reality, the above calculated the van 't Hoff factor times the analytical concentration. When 0.5000 g of this mixture is dissolved in enough water to form 1.000 L of solution, the osmotic pressure at 25.0 ☌ is observed to be 0.3990 atm. Problem #3: A solid mixture contains MgCl 2 and NaCl. We are now ready to calculate the percent ionization: (If you're not sure where the x, x and 0.035 - x came from, write down the dissociation equation for HNO 2.) Now, we need to add up the molarities of the three solute particles to equal 0.0384. Next, consider the dissociation equation: This tells use that the true concentration (when all three types of non-water particles are added up) is 0.0384 M. Calculate the percent ionization of the acid.Ġ.93 atm = i (0.035 mol L¯ 1) (0.08206 L atm mol¯ 1 K¯ 1) (295 K) I'll leave you to check my answer.Ī 0.035 M aqueous nitrous acid (HNO 2) solution has an osmotic pressure of 0.93 atm at 22.0 ☌. Now we have all the data to solve the last step, plugging into the K a expression and solving it. Let us call that value 'y.' Therefore the is 0.010 - y and here comes a key statement:
#OSMOTIC PREASSURE MOLAR MASS FINDER HOW TO#
So then the question becomes, how to figure out, and ? For this, we turn to the equation of dissociation: The next step is to ponder the K a expression:įor me, the key is to see that, since the K a value is the unknown, ALL the other values must be known. When all the individual non-water particles are counted up, it all comes out to 0.0117 M. 0.010 mole of HOCN was put into the solution, Some, not all, of the molecules fell apart into H 3O + and OCN¯. In fact, they did dissociate, enough so that the solution is acting like it is 0.0117 M and not 0.010 M. The 1.17 tells us that the HOCN molecules we put in (call it 100%) is actually creating an osmotic pressure that would require 117% of what we put in, IF the molecules did not dissociate. However, in the solution it is a different story and i = 1.17 is what tells us that. If we evaporated all the water, we would recover 0.010 mol of HOCN. What the person created is called the analytical concentration. The person was very careful at each step, used appropriate glassware that was carefully cleaned, etc, etc. Then they dissolved it in some distilled water and added sufficient water to then make 1.0 L of the solution. Someone weighs out the necessary grams of HOCN to equal 0.010 mole of it. Imagine, if you will the making of 0.010 M HOCN. Keep in mind that i is associated with the degree of dissociation. I want to pause for a moment and ponder the physical meaning of i and exactly what 1.17 means. Notice I've included a conversion of torr to atm. Calculate the K a for HOCN.įor what is given and what is not given in the problem.ģ) a constant (R) = 0.08206 L atm mol¯ 1 K¯ 1 The osmotic pressure of a 1.0 x 10¯ 2 M solution of cyanic acid (HOCN) is 217.2 torr at 25 ☌. Worksheet - Osmosis Problems - AP level Worksheet - Osmosis Problems - AP level
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