The Osmotic pressure at STP of a solution made from 1 L of NaCl (aq) containing 117 g of NaCl is: a) 44.77 atm b) 44.87 atm c) 89.54 atm d) 117 atm Hint: the tricky part has to do with the moles of NaCl

People, the solution is simple.... Osmotic Pressure=MRTi THis can be manipulated to Osmotic Pressure=nRTi/V n= 117/58.5= 2 moles NaCl R*T at STP= 22.4 i= 2, because NaCl breaks up into two ions Therefore Osmotic pressure= (2)(22.4)(2)= 89.6

Thats what I thought as 2 moles but it is actually 4 moles of NaCl and I am trying to figure it out why it is 4 moles. The rest is just plugging in the numbers to get the answer.

It is 2 moles of NaCl, but NaCl completely dissociates in water so you get two moles of Na+ and 2 moles of Cl- total. If it didn't dissociate you would have: (2)(22.4)(1) = 44.8 According to the equation OP = nRTi/V, but instead you get: (2)(22.4)(2) = 89.6

Osmotic pressure, Boiling-Point Elevation, and Freezing-Point Depression all depend on Colligative Properties. So for this example, NaCl, each mole dissociates into 2 ions (Na+ and Cl-). If you got something that doesn't dissociate, like an organic compound (i.e. ethanol, C2H5OH) you only have the one mole. If you get a problem that deals with Colligative properties, you'll have to use this approach.