MEDIUMMCQ SINGLEJEE Advanced ChemistryPhysical ChemistrySolutionsRelative lowering of vapour pressure
The vapour pressure of pure solvent is 0.8 mm of Hg at a particular temperature. On addition of a non-volatile solute ‘A’ the vapour pressure of solution becomes 0.6 mm of Hg. The mole fraction of component ‘A’ is
A.0.25✓ Correct
B.0.75
C.0.5
D.0.35
Explanation
According to Raoult's law, the lowering in vapour pressure is written as, 0 2 0 P P X P We are provided with the following data: Vapour pressure of the pure solvent is 0P 0.8 mm of Hg Vapour pressure of solvent on the addition of solute is P 0.6 mm of Hg Let's substitute the values in the equation. We have, 2 2 0.8 0.6 X 0.8 X 0.25 >> Thus, (a) is the correct option.
Practice More Like This
Take chapter-wise tests, track your progress, and master Solutions.
Start Free Practice →