72 511
Assignments Done
98,8%
Successfully Done
In April 2019

Answer to Question #17435 in Physical Chemistry for skand

Question #17435
I have a question in colligative properties that if there are compunds like NaCl , C6H12O6(glucose) , CaCl2(calcium cholride) of the same concentration and in aqueous solution how to find which has more lowering in vapour pressure ? Pls help....
Expert's answer
The osmotic pressure Π of an ideal solution with low concentration can be approximated using the Morse equation:

П = iMRT

i is the dimensionless van 't Hoff factor
M is the molarity
R = 8.3145 J K-1 mol-1 (5 s.f.) is the gas constant
T is the thermodynamic (absolute) temperature

The van 't Hoff factor (named after J. H. van 't Hoff) is a measure of the effect of a solute upon colligative properties such as osmotic pressure, relative lowering in vapor pressure, elevation of boiling point and freezing point depression. The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved, and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van' t Hoff factor is essentially 1. For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance. This is true for ideal solutions only, as occasionally ion pairing occurs in solution. At a given instant a small percentage of the ions are paired and count as a single particle. Ion pairing occurs to some extent in all electrolyte solutions. This causes deviation from the van 't Hoff factor. The deviation for the van 't Hoff factor tends to be greatest where the ions have multiple charges. So for C6H12O6 (glucose, non-electrolyte) i=1; for NaCl i = 2 (NaCl <-> Na+ + Cl-); for CaCl2 i = 3 (CaCl2 <-> Ca2+ + 2Cl-).

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question

Submit
Privacy policy Terms and Conditions