(i) Derive relationship between relative lowering of vapour pressure and molar mass of non volatile solute. (i

Question 97 of 97

Q. (i) Derive relationship between relative lowering of vapour pressure and molar mass of non volatile solute.

(ii) Write statement of second law of thermodynamics.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026Subjective· 3mImportance★★★★★

100% · 97/97 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Relative Lowering of Vapour Pressure

From Intuition to a Precise Law

Imagine a beaker of pure water left open. Water molecules at the surface are constantly escaping into the air above — that's evaporation. The pressure exerted by those vapour molecules when the system reaches equilibrium is the vapour pressure of pure water.

Now dissolve some sugar in that water. The sugar molecules are non-volatile — they don't evaporate. They sit at the surface, taking up space. Fewer water molecules are now at the surface to escape into the vapour phase. The result? The vapour pressure above the solution is lower than that above pure water.

That's the intuition: a non-volatile solute physically blocks some solvent molecules from leaving the liquid, so fewer vapour molecules form above the solution.

The Precise Statement

The relative lowering of vapour pressure is defined as:

\frac{P^0 - P}{P^0}

where $P^0$ is the vapour pressure of the pure solvent and $P$ is the vapour pressure of the solution.

\frac{P^0 - P}{P^0} = x_{\text{solute}}

Here $x_{\text{solute}}$ is the mole fraction of the non-volatile solute in the solution.

This is Raoult's law for a non-volatile solute. The law says: the fractional decrease in vapour pressure depends only on how many solute particles are present, not on what they are. That's what makes it a colligative property — it depends on the number of solute particles, not their identity.

Why "Relative" and Why "Lowering"?

The word relative is crucial. The absolute drop in pressure ( $P^0 - P$ ) depends on the solvent itself — water and ethanol have very different $P^0$ values. But the fraction of that drop, relative to the pure solvent's pressure, is the same for the same mole fraction of solute, regardless of the solvent.

The lowering is simply $P^0 - P$ , the amount by which the vapour pressure has fallen.

Tip

A quick way to remember: if you add a non-volatile solute, the vapour pressure always goes down. The relative lowering tells you how much it went down as a fraction of the original.

A Simple Example

Suppose you dissolve glucose in water such that the mole fraction of glucose is 0.05. The vapour pressure of pure water at that temperature is, say, 23.8 mm Hg.

Then:

\frac{P^0 - P}{P^0} = 0.05

So:

P^0 - P = 0.05 \times 23.8 = 1.19 \text{ mm Hg}

And the vapour pressure of the solution is:

P = 23.8 - 1.19 = 22.61 \text{ mm Hg}

The relative lowering is 0.05 — a pure number, independent of the units of pressure.

--- …

Start your 14-day free trial to unlock the full solution →

Raoult's law gives relative lowering of vapour pressure = mole fraction of solute, which is used to find the solute's molar mass; the second law is stated via the Clausius/entropy statement.

(i) Relative lowering of vapour pressure and molar mass:

For a dilute solution of a non-volatile solute in a volatile solvent, by Raoult's law the vapour pressure of solution $p$ is proportional to the mole fraction of solvent $x_1$ : $p = p^{\circ}x_1$ , where $p^{\circ}$ is the vapour pressure of pure solvent.

Since $x_1 + x_2 = 1$ (solute mole fraction $x_2$ ):

$p^{\circ} - p = p^{\circ}x_2 \Rightarrow \dfrac{p^{\circ}-p}{p^{\circ}} = x_2$

This ratio is called the relative lowering of vapour pressure, and it equals the mole fraction of solute.

For a solution with $w_1$ g of solvent (molar mass $M_1$ ) and $w_2$ g of solute (molar mass $M_2$ ), and for a dilute solution ( $n_1 \gg n_2$ ):

$x_2 = \dfrac{n_2}{n_1+n_2} \approx \dfrac{n_2}{n_1} = \dfrac{w_2/M_2}{w_1/M_1} = \dfrac{w_2 M_1}{w_1 M_2}$

So: $\dfrac{p^{\circ}-p}{p^{\circ}} = \dfrac{w_2 M_1}{w_1 M_2}$ , giving $M_2 = \dfrac{w_2 M_1 p^{\circ}}{w_1 (p^{\circ}-p)}$

This relation allows the molar mass $M_2$ of an unknown non-volatile solute to be determined experimentally by measuring the relative lowering of vapour pressure.

…

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set A1 markMCQ
$Q. Aqueous solution of which of the following will have lowest vapour pressure ?$
(a) 0.1 M BaCl2
(b) 0.1 M Urea
(c) 0.1 M Na2SO4
(d) 0.1 M Na3PO4
›Reveal solutionSolution
Relative lowering of vapour pressure depends on the number of dissolved particles; Na3PO4 dissociates into 4 ions, the most among the options.

Lowering of vapour pressure is a colligative property, proportional to the total particle concentration (i x molarity). At 0.1 M:

Urea: i = 1 → 0.1

BaCl2: i = 3 → 0.3

Na2SO4: i = 3 → 0.3 …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. 12 g of urea is dissolved in 1 L of water and 68.4 g of sucrose is dissolved in 1 L of water. Relative lowering of vapour pressure of urea solution is-$
(a)
(i) Greater than sucrose solution
(b)
(ii) Less than sucrose solution
(c)
(iii) Double that of sucrose solution
(d)
(iv) Equal to that of sucrose solution
›Reveal solutionSolution
Both solutions contain the same number of moles of solute (0.2 mol) in the same amount of water, so their mole fractions of solute — and hence their relative lowering of vapour pressure — are equal. Correct option: (iv).

Concept. Relative lowering of vapour pressure (RLVP) is a colligative property: it depends on the number of solute particles, not their nature. By Raoult's law for a non-volatile solute,

$\frac{p^{0}-p_s}{p^{0}} = x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}}+n_{\text{solvent}}}$

Why compare moles. Since both solutions use the same solvent quantity (1 L water), the one with more solute moles has the larger RLVP. So we just compare solute moles.

Steps.

Urea: $n = \dfrac{12}{60} = 0.2\ \text{mol}$ (molar mass of urea, $\text{NH}_2\text{CONH}_2 = 60\ \text{g mol}^{-1}$ ).

Sucrose: $n = \dfrac{68.4}{342} = 0.2\ \text{mol}$ (molar mass of sucrose, $\text{C}_{12}\text{H}_{22}\text{O}_{11} = 342\ \text{g mol}^{-1}$ ). …
Start your 14-day free trial to see the full solution →
CBSE 2025Set D1 markMCQ
$Q. Which of the following is not a colligative property?$
(a) Vapour pressure
(b) Depression of freezing point
(c) Osmotic pressure
(d) Elevation of boiling point
›Reveal solutionSolution
Vapour pressure is not a colligative property; the colligative one is the RELATIVE LOWERING of vapour pressure.

A colligative property depends only on the number of solute particles present, not on their nature. The four colligative properties are:

Relative lowering of vapour pressure

Depression in freezing point

Elevation in boiling point

Osmotic pressure …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Dissolution of Solute in Solvent:$
(a) Lowers its vapour pressure
(b) Increases its vapour pressure
(c) Increases its freezing point
(d) Lowers its boiling point
›Reveal solutionSolution

Adding a non-volatile solute to a solvent reduces the fraction of solvent molecules at the surface that can escape into vapour, so the vapour pressure of the solution is lower than that of the pure solvent.

According to Raoult's law, for a solution of a non-volatile solute in a volatile solvent, the vapour pressure of the solution (p) is p = p° × x_solvent, where p° is the vapour pressure of pure solvent and x_solvent < 1 is the mole fraction of solvent (since some of the moles are now solute). Because x_solvent is less than 1, p < p°, i.e. the vapour pressure is lowered on dissolving the solute.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following is not a colligative property?$
(a) relative lowering of vapour pressure
(b) elevation in boiling point
(c) depression in freezing point
(d) depression in boiling point
›Reveal solutionSolution

Colligative properties depend only on the NUMBER of solute particles, not their identity: relative lowering of vapour pressure, elevation in boiling point, depression in freezing point, and osmotic pressure. Boiling point never falls when a non-volatile solute is added, so 'depression in boiling point' is not a real colligative property.

When a non-volatile solute is dissolved in a solvent, the vapour pressure of the solution is lowered (Raoult's law). A lower vapour pressure means the solution needs to be heated to a HIGHER temperature to reach atmospheric pressure and boil, so the boiling point RISES (elevation), never falls.

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set 56/2/11 markMCQ
$Q.The relative lowering of vapour pressure of an aqueous solution containing non-volatile solute is 0·0225. The mole fraction of the non-volatile solute is : (A) 0·80 (B) 0·725 (C) 0·15 (D) 0·0225$
›Reveal solutionSolution
The relative lowering of vapour pressure equals the mole fraction of the solute in the solution. Given the lowering is 0.0225, the mole fraction of the non-volatile solute is directly 0.0225, which corresponds to option (D).

Why This Works: The Concept of Relative Lowering

When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure of the solvent above the solution is lower than that above the pure solvent. This is a colligative property — it depends only on the number of solute particles, not their identity.

The key relationship, known as Raoult’s law for a non-volatile solute, states:
$\frac{P^0 - P}{P^0} = x_{\text{solute}}$
Here:

$P^0$ = vapour pressure of pure solvent

$P$ = vapour pressure of the solution

$\frac{P^0 - P}{P^0}$ = relative lowering of vapour pressure

$x_{\text{solute}}$ = mole fraction of the non-volatile solute in the solution

The beauty of this formula is its directness: the relative lowering is exactly equal to the mole fraction of the solute. No extra constants, no temperature dependence (as long as the solvent is the same). This is because the solute molecules occupy some of the surface sites, reducing the number of solvent molecules that can escape into the vapour phase.

Watch out

A common mistake is to confuse the relative lowering with the mole fraction of the solvent. Remember: the lowering is proportional to the solute's mole fraction, not the solvent's. If you ever feel unsure, recall that for a pure solvent (no solute), the lowering is zero, and the solute mole fraction is zero — they match.

Step-by-Step Solution

Identify the given data. The problem states: "The relative lowering of vapour pressure of an aqueous solution containing non-volatile solute is 0.0225." So, $\frac{P^0 - P}{P^0} = 0.0225$ . …
Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. According to Raoult's law, the relative lowering of vapour pressure is equal to -$
(a) Molarity of the solution
(b) Molality of the solution
(c) Mole fraction of the solute
(d) Mole fraction of the solvent
›Reveal solutionSolution

Raoult's law states directly that the relative lowering of vapour pressure equals the solute's mole fraction.

According to Raoult's law, for a solution of a non-volatile solute in a volatile solvent, the relative lowering of vapour pressure is given by:

$\frac{p°_{solvent} - p_{solution}}{p°_{solvent}} = x_{solute}$

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. Relative lowering in vapour pressure is equal to$
(a) molarity of solution
(b) molality of solution
(c) mole fraction of solute
(d) mole fraction of solvent
›Reveal solutionSolution

Raoult's law directly equates the relative lowering of vapour pressure to the solute's mole fraction.

For a solution of a non-volatile solute in a volatile solvent, Raoult's law states:

(P0 - Ps)/P0 = x(solute) …

Start your 14-day free trial to see the full solution →
CBSE 2017Set ANNUAL1 mark
$Q. Why does vapour pressure of a liquid decrease when a non-volatile solute is added into it ?$
›Reveal solutionSolution

A non-volatile solute reduces the fraction of solvent molecules at the surface, lowering the rate of escape into vapour and hence the vapour pressure.

Vapour pressure arises from solvent molecules escaping from the liquid surface into the vapour phase. When a non-volatile solute is dissolved, some solute particles occupy positions at (or near) the surface, effectively reducing the fraction/number of solvent molecules present per unit area of the surface. This lowers the rate at which solvent molecules can escape into the vapour phase, so a smaller vapour pressure is need …

Start your 14-day free trial to see the full solution →

←Previous questionQ27 — Calculate the osmotic pressure of solution containing 0.822 gm of sucr…→Next chapterCh 3Ionic Equilibria

Q.(i) Derive relationship between relative lowering of vapour pressure and molar mass of non volatile solute.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days