Chemistry · Solutions · NEET
Start from (p° - p)/p° = x₂ = mole fraction of solute. Write x₂ = n₂ / (n₁ + n₂). For a dilute solution n₂ is very small, so n₁ + n₂ ≈ n₁, giving (p° - p)/p° ≈ n₂/n₁ = (w₂/M₂) / (w₁/M₁). Rearranged: M₂ = (w₂ × M₁ × p°) / (w₁ × (p° - p)). Here w₂, M₂ are the mass and molar mass of solute; w₁, M₁ are the mass and molar mass of solvent (water = 18).
Moles = mass ÷ molar mass. The solute mass w₂ is given but its molar mass M₂ is the unknown we want. So we write moles of solute as w₂/M₂. For the solvent both mass w₁ and molar mass M₁ are known (for water M₁ = 18), so n₁ = w₁/M₁ is a number you can compute. This turns the mole-fraction equation into one equation with one unknown, M₂.
Use the simple version only for dilute solutions, where the number of solute moles n₂ is tiny compared to solvent moles n₁. Then n₁ + n₂ ≈ n₁ and you may drop n₂ from the denominator. NEET numerical problems are almost always dilute, so this approximation is expected. If the solution is concentrated, keep the full form x₂ = n₂/(n₁ + n₂).
p° is the vapour pressure of the PURE solvent, and p is the vapour pressure of the solution (after solute is added). p is always less than p°. A common NEET trick: at the normal boiling point of water (100 °C) the vapour pressure of pure water equals 760 mm Hg, so if the problem is at 100 °C, use p° = 760 mm even if it is not stated.
No. Relative lowering of vapour pressure is a colligative property. It depends only on the number of solute particles, not on their identity or size. Sugar and urea at the same mole fraction lower the vapour pressure by the same amount. This is exactly why the formula lets us find molar mass from a physical measurement.
18 is the molar mass of water, so it is the SOLVENT term M₁. Students often wrongly put 18 in place of the unknown solute molar mass. Remember: water is the solvent here, so n₁ = w₁/18. The solute's molar mass M₂ is what you are solving for and it will usually come out to some other value.
At 100 °C the vapour pressure of a solution of 6.5 g of a solute in 100 g of water is 732 mm. If Kb = 0.52, the boiling point of this solution will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the drop in vapour pressure divided by the pure solvent's vapour pressure, written (p° - p)/p°. Raoult showed it equals the mole fraction of the non-volatile solute. It is a pure number (no units) and depends only on how many solute particles are present.
You need four things: the mass of the solute (w₂), the mass of the solvent (w₁), the molar mass of the solvent (M₁, e.g. 18 for water), and the vapour pressures p° (pure solvent) and p (solution). Then M₂ = (w₂ × M₁ × p°) / (w₁ × (p° - p)).
Yes. It is one of the four colligative properties, along with elevation of boiling point, depression of freezing point, and osmotic pressure. All four depend only on the number of solute particles, which is why each can be used to measure molar mass.
NEET almost every year asks a numerical from Solutions where you find molar mass or link it to boiling point / freezing point. The 2016 PYQ above literally uses vapour pressure lowering as the first step. Mastering this formula unlocks the whole colligative-properties chain of questions.
