Dissolving a solute into a solvent has four colligative effects on the solution:
ID: 880563 • Letter: D
Question
Dissolving a solute into a solvent has four colligative effects on the solution:
1) vapor pressure lowering – the presence of dissolved solute lowers the vapor pressure of a solution via Raoult’s Law:
(Vapor Pressure of Solution) = (Mole Fraction of Solvent) • (Vapor Pressure of Pure Solvent)
or
Psolution = Xsolvent• Psolvent
This lowering of vapor pressure due to the presence of dissolved solute is given as:
P = Xsolute• Psolvent• i
where “i” is the van’t Hoff factor, which is, ideally, the number of particles that result in the dissolution of one particle of solute. That is, if 1 molecule of sucrose dissolves, i = 1 because sucrose does not dissociate into ions. But if one formula unit of NaCl dissolves, i = 2 because, ideally, 2 ions dissociate from each NaCl in aqueous solution. Likewise, MgCl2 should yield three ions, so that i = 3 ideally. (I refer to “ideally” because just as an ideal solution exhibits Hsolution = 0, an ideal solution also obeys Raoult’s Law above, which requires that van’t Hoff factors are whole numbers from 1 up. Of course, most solutions are not 100% ideal, but for the intents and purposes of this course, we assume ideal solutions. The P needs to be subtracted from the vapor pressure of the pure solvent to get the vapor pressure of the solution.
2) due to the colligative effect of vapor pressure lowering, we also see boiling point elevation with the presence of a dissolved solute, given as:
Tb = kb• m • i
where m is molality, i is van’t Hoff factor, and kb is the boiling-point elevation constant specific to the solvent being considered (this number will always be provided to you). The Tb needs to be added to the boiling point of the pure solvent to get the boiling point of the solution.
3) freezing point (or melting point) depression:
Tf = kf• m • i
where m is molality, i is van’t Hoff factor, and kf is the freezing-point depression constant specific to the solvent being considered (this number will always be provided to you). The Tf needs to be subtracted from the freezing point of the pure solvent to get the freezing point of the solution.
4) osmotic pressure development:
= M•R•T•i
where is osmotic pressure (in atm), M = molarity, R = universal gas constant 0.0821 atm•lit / mol•K, T = absolute temperature in Kelvin, and i is van’t Hoff factor.
What is the vapor pressure of a solution prepared by adding 50.0 g of NaCl to 250. g of pure water at 25oC? (Water has a vapor pressure of 0.036atm at 25oC)
A. 0.006 atm
B. 0.030 atm
C. 0.032 atm
D. 0.002 atm
E. 0.034 atm
F. 0.004 atm
Explanation / Answer
the vapou pressure will decrease by addition of solvent by 0.006atm. Hence, new vapour presuure of the solution is 0.036atm-0.006atm which equals 0.030atm (ANSWER)
answer is B. 0.030atm