Question

10mL of liquid ‘A’ is mixed with 10mL of liquid ‘B’, the volume of the resultant solution is 19.9mL. What type of deviation expected from Raoult’s law?

Answer 1

When the resultant volume of the solution is less than the sum of the volume of solutions, then the deviations from Raoult’s law is negative and when the vapour pressure is higher than the solution than it positive deviation

Complete step by step answer:
According to Raoult’s Law definition, it states the solvent is particle vapor pressure in a solution is equal or identical to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.
Equation of Raoult’s law is.
${P_{Solution}} = {X_{Solvent}}P_{Solvent}^0$
Where each symbol represents,
${P_{Solution}} \to$ Vapour pressure of Solution
${X_{Solvent}} \to$ mole fraction of solvent
$P_{Salvent}^0 \to$ Vapour pressure of pure solvent.
In the given question, the volume of resultant solution is given to be 19.9 ml.
But in actual,
The sum of the volume of two solutions is 10 ml (Volume of solution A) + 10 ml (Volume of solution B) = 20 ml (Volume of both solution)
As you can infer,
The volume of resultant solution is less the volume of resultant solution is less than [Volume of solution A + Volume of Solution B]
So, it can be said that when two liquids are mixed, there is a decrease in volume.
The decrease in volume results in the negative deviation.
To conclude, negative deviation is expected from Raoult's Law.

Note:
Raoult’s Law is basically apt for describing ideal solution but as you know ideal solution but as you know ideal solutions are hard to find and they are rare.
Moreover, many of the liquids are in mixture and they do not have the same uniformity and they tend to deviate away from law.
Always Raolt’s law works well for the ideal solution. Since, the ideal solutions are rare, it seems to be the limitation for Raoult’s law.