Question
Question: The vapor pressure of pure water at \(75^\circ\)is \(296torr\) then the vapour pressure of lowering ...
The vapor pressure of pure water at 75∘is 296torr then the vapour pressure of lowering due to 0.1m solute is:
A.0.533torr
B.0.296torr
C.0.333torr
D.0.428torr
Solution
Raoult’s law states that partial vapour pressure of a solvent in a solution or in mixture is equal or to the vapour pressure of the pure solvent which is multiplied by its mole fraction in the solution. Raoult's law is useful in describing ideal solution
Complete step by step solution:
We know that,
Raoult’s law is given as,
Psolvent=Xsolution×PSolvent0 ……(1)
Here,
Psolvent= vapour pressure of solvent
Xsolution= mole fraction of the solvent
PSolvent0= vapour pressure of solvent
It is given to us that,
Vapor pressure of pure water = 296torr
Temperature = 75∘C
Amount of Solute = 0.1m
Number of moles of water = 181000
Using the given values in equation (1), we will now calculate the value of Psolvent
Therefore, the correct option is A.
Additional information:
Few limitations to Raoult’s law are:
1)Raoult's law is useful for describing ideal solutions. However, ideal solutions are rare and hard to find.
2)The negative deviation is seen when the vapour pressure is lower than that of the expected from Raoult's law. A positive deviation is seen when the cohesion between the similar molecules is greater or exceeds the adhesion between dissimilar molecules. Both components of the mixture can easily escape from the solution.
Note: Raoult's law is valid only in ideal solutions. In an ideal solution, the solvent-solute interaction is the same as that of the solvent - solvent or solute - solute interaction. This tells that both the solute and the solvent take the same amount of energy in order to escape to the vapour phase when they exist in their pure states.