Chemistry · Solutions · NEET
It is a ratio, not a pressure. Take the drop in vapour pressure (p° − p) and divide it by the pure solvent's vapour pressure (p°). So relative lowering = (p° − p) / p°. Because it is a ratio, it has no units. NEET loves this because Raoult's law says this ratio equals the mole fraction of the solute.
At the liquid surface, only solvent molecules can escape into vapour (the solute is non-volatile, so it stays). When solute particles sit at the surface, they take up space and fewer solvent molecules are able to leave. Fewer escaping molecules means lower vapour pressure. This is why it depends only on the NUMBER of solute particles, making it a colligative property.
Solute. This is the most common NEET slip. (p° − p) / p° = x(solute) = n(solute) / [n(solute) + n(solvent)]. The vapour pressure p itself equals p° × x(solvent). So remember: p = p° × x(solvent), but the LOWERING equals x(solute).
Only for dilute solutions. In a dilute solution the solute moles are very small compared to solvent moles, so n(solute)+n(solvent) ≈ n(solvent). Then x(solute) ≈ n(solute)/n(solvent). NCERT uses this shortcut to find molar mass. For concentrated solutions, use the full formula with both terms in the denominator.
p° (p-naught) is the vapour pressure of the PURE solvent alone. p is the vapour pressure of the SOLUTION after adding the solute. p is always smaller than p° because the solute lowers it. The drop (p° − p) is the 'lowering'.
No. It depends only on how many solute particles are present, not on their chemical nature. Sugar, urea, or glucose all give the same lowering if the number of particles is the same. That is why it is called a colligative property. (Electrolytes like NaCl split into ions, so they give MORE particles — that needs the van't Hoff factor.)
At 100 °C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be:
The molality of 2.5 g of ethanoic acid (Molar mass: 60 g/mol) in 75 g of benzene is (part of a multi-statement question that uses concentration and vapour-pressure ideas):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
(p° − p) / p° = x(solute), where p° is the vapour pressure of the pure solvent, p is the vapour pressure of the solution, and x(solute) is the mole fraction of the non-volatile solute.
Because its value depends only on the number of solute particles dissolved, not on their chemical identity. More particles cause a bigger lowering.
Rearrange (p° − p)/p° = (w/M)/(W/Mo) for dilute solutions, where w and M are the mass and molar mass of solute, and W and Mo are the mass and molar mass of solvent. Solve for M. This is a standard NEET calculation.
p° is the vapour pressure of the pure solvent. p is the lower vapour pressure of the solution after adding a non-volatile solute. p is always less than p°.
Yes. More solute particles means a higher mole fraction of solute, so the relative lowering (p° − p)/p° increases and p drops further.
