Using Raoult's law explain how the total vapour pressure over the solution is related to mole fraction of comp

NCERT Exemplar · Q56

Q.Using Raoult's law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.

(i)

CHCl_3(l)

and

CH_2Cl_2(l)

(ii)

NaCl(s)

and

H_2O(l)

Sikkim CbseLong· 3mImportance★★★★★

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Concept understanding — Raoult's Law

Raoult's Law: From Intuition to Precision

Imagine you have a beaker of pure water at room temperature. Some water molecules at the surface have enough energy to escape into the air above — that's vapour pressure. Now dissolve some sugar in that water. The sugar molecules take up space at the surface, blocking some water molecules from escaping. Fewer water molecules can leave the liquid per second, so the vapour pressure drops.

That's the core intuition: a non-volatile solute lowers the solvent's vapour pressure simply by getting in the way.

But what if both components can evaporate — say, a mixture of benzene and toluene? Then both kinds of molecules crowd the surface, and both contribute to the total vapour pressure. The question becomes: how much does each contribute?

The Precise Statement

For a solution of volatile liquids, Raoult's Law says:

p_i = x_i \, p_i^*

where:

$p_i$ = partial vapour pressure of component $i$ above the solution
$x_i$ = mole fraction of component $i$ in the liquid solution
$p_i^*$ = vapour pressure of pure component $i$ at the same temperature

The law applies to each volatile component separately. The total vapour pressure above the solution is simply the sum:

P_{\text{total}} = p_1 + p_2 = x_1 p_1^* + x_2 p_2^*

What This Means Physically

The mole fraction $x_i$ tells you the fraction of molecules at the surface that are of type $i$ . If half the molecules in the liquid are benzene ( $x_{\text{benzene}} = 0.5$ ), then roughly half the surface sites are occupied by benzene molecules. So the rate at which benzene escapes should be about half the rate from pure benzene — hence $p_{\text{benzene}} = 0.5 \times p_{\text{benzene}}^*$ .

This is a linear relationship: plot $p_i$ against $x_i$ , and you get a straight line from the origin (when $x_i = 0$ , $p_i = 0$ ) up to $p_i^*$ (when $x_i = 1$ , pure component).

Tip

Raoult's Law is an idealisation. It works best when the two liquids are chemically similar — same type of intermolecular forces (e.g., both non-polar, or both with similar hydrogen bonding). Benzene–toluene is a classic example. When the molecules interact very differently (like ethanol and water), the law fails — that's when you get deviations from Raoult's Law.

A Concrete Example

Suppose you mix 2 moles of benzene ( $p^* = 100$ mm Hg) with 3 moles of toluene ( $p^* = 40$ mm Hg) at 25°C.

Mole fractions:

$x_{\text{benzene}} = \frac{2}{2+3} = 0.4$
$x_{\text{toluene}} = \frac{3}{5} = 0.6$

Partial pressures:

$p_{\text{benzene}} = 0.4 \times 100 = 40$ mm Hg
$p_{\text{toluene}} = 0.6 \times 40 = 24$ mm Hg

Total vapour pressure: $40 + 24 = 64$ mm Hg

Notice: the total pressure is not a simple average of the pure pressures. It's a weighted average, with mole fractions as weights.

Why This Matters …

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Raoult’s law relates the partial vapour pressure of each component to its mole fraction in the liquid phase. For ideal solutions (like chloroform–dichloromethane), total pressure varies linearly with composition; for non‑volatile solutes (like NaCl in water), the total pressure is simply the vapour pressure of the solvent, lowered by the solute.

The core idea

Raoult’s law says: At a given temperature, the partial vapour pressure of a component in a liquid solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the liquid mixture.

p_i = x_i \, p_i^\circ

where $p_i$ is the partial pressure of component $i$ above the solution, $x_i$ is its mole fraction in the liquid, and $p_i^\circ$ is the vapour pressure of pure $i$ at that temperature.

The total vapour pressure above the solution is the sum of all partial pressures:

P_{\text{total}} = \sum_i p_i = \sum_i x_i p_i^\circ

How this total pressure changes with composition depends entirely on whether both components are volatile and whether the solution is ideal or not.

(i) $CHCl_3(l)$ and $CH_2Cl_2(l)$ — two volatile liquids, nearly ideal

Chloroform ( $CHCl_3$ ) and dichloromethane ( $CH_2Cl_2$ ) are both volatile organic liquids. Their molecular structures are similar, and they mix without strong interactions — so the solution behaves very close to an ideal solution.

Both components contribute to vapour pressure. Let $A = CHCl_3$ and $B = CH_2Cl_2$ .

p_A = x_A p_A^\circ \quad \text{and} \quad p_B = x_B p_B^\circ

Total pressure is a linear function of mole fraction. Since $x_B = 1 - x_A$ , we have:

P_{\text{total}} = x_A p_A^\circ + (1 - x_A) p_B^\circ = p_B^\circ + x_A (p_A^\circ - p_B^\circ)

This is a straight line when plotted against $x_A$ (or $x_B$ ).

At $x_A = 0$ , $P_{\text{total}} = p_B^\circ$ ; at $x_A = 1$ , $P_{\text{total}} = p_A^\circ$ .

The vapour composition is different from the liquid composition (that’s the basis of fractional distillation), but the total pressure follows Raoult’s law exactly.

Tip

For an ideal binary solution of two volatile liquids, the total vapour pressure always lies between the two pure vapour pressures. If $p_A^\circ > p_B^\circ$ , then $P_{\text{total}}$ increases linearly as $x_A$ increases.

(ii) $NaCl(s)$ and $H_2O(l)$ — a non‑volatile solute in a volatile solvent

Here, NaCl is a solid salt that dissolves in water. The key difference: NaCl has essentially zero vapour pressure — it does not evaporate. Only water contributes to the vapour above the solution.

Only the solvent is volatile. Let $A = H_2O$ (solvent) and $B = NaCl$ (solute).

p_{\text{NaCl}} \approx 0 \quad \Rightarrow \quad P_{\text{total}} = p_{\text{water}}

Raoult’s law applies to the solvent. For the solvent:

p_{\text{water}} = x_{\text{water}} \, p_{\text{water}}^\circ

Since $x_{\text{water}} < 1$ (because NaCl is present), the vapour pressure of water above the solution is lower than that of pure water. …

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Method: Raoult's Law for Vapour Pressure of Solutions

Raoult's Law states that for a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the solution:

p_i = p_i^0 \cdot x_i

Where:

$p_i$ = partial vapour pressure of component $i$ above the solution
$p_i^0$ = vapour pressure of pure component $i$ at the same temperature
$x_i$ = mole fraction of component $i$ in the liquid solution

Total vapour pressure above the solution is the sum of partial pressures:

P_{\text{total}} = p_1 + p_2 = p_1^0 x_1 + p_2^0 x_2

(i) $CHCl_3(l)$ and $CH_2Cl_2(l)$ — Ideal Solution

Step 1: Identify the nature of the solution

Both are volatile liquids with similar molecular structures (chlorinated hydrocarbons). They mix without significant interaction changes → ideal solution.

Step 2: Apply Raoult's law for both components

Let $A = CHCl_3$ and $B = CH_2Cl_2$

p_A = p_A^0 x_A \quad \text{and} \quad p_B = p_B^0 x_B

Step 3: Write total vapour pressure

Since $x_B = 1 - x_A$ :

P_{\text{total}} = p_A^0 x_A + p_B^0 (1 - x_A)

Step 4: Interpret the relationship

$P_{\text{total}}$ varies linearly with mole fraction $x_A$
The graph of $P_{\text{total}}$ vs $x_A$ is a straight line between $p_B^0$ (at $x_A = 0$ ) and $p_A^0$ (at $x_A = 1$ )

Key result: For ideal solutions, total vapour pressure is a linear function of mole fraction.

(ii) $NaCl(s)$ and $H_2O(l)$ — Non-volatile Solute in Volatile Solvent

Step 1: Identify the nature

$NaCl$ is a non-volatile solid (its vapour pressure ≈ 0)
$H_2O$ is the volatile solvent

Step 2: Apply Raoult's law only to the volatile component

Let solvent = $H_2O$ (component 1), solute = $NaCl$ (component 2, $p_2^0 \approx 0$ )

p_{\text{solvent}} = p_{\text{solvent}}^0 \cdot x_{\text{solvent}}

Step 3: Express mole fraction of solvent

If $n_1$ = moles of water and $n_2$ = moles of $NaCl$ :

x_{\text{solvent}} = \frac{n_1}{n_1 + n_2}

Step 4: Write total vapour pressure …

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Common Mistakes with Henry’s Law & Raoult’s Law (and How to Avoid Them)

Mistake 1: Confusing Henry’s Law with Raoult’s Law

The error: Students often apply Henry’s Law ( $p = K_H x$ ) to all solutions, even when the solute and solvent are chemically similar (like $CHCl_3$ and $CH_2Cl_2$ ). They forget that Raoult’s Law applies only when intermolecular forces between components are nearly identical.

How to avoid:

Raoult’s Law → $p_i = x_i p_i^0$ (for ideal solutions or solvent in dilute solutions)
Henry’s Law → $p_i = K_H x_i$ (for solute in very dilute solutions, where solute–solvent forces differ)
Key test: If the two liquids are structurally similar (both chlorinated hydrocarbons), expect Raoult’s Law behaviour. If one is a gas or a solid dissolving in a liquid, use Henry’s Law for the solute.

Mistake 2: Applying Raoult’s Law to a Non-Volatile Solute (like NaCl)

The error: Students write $p_{NaCl} = x_{NaCl} p_{NaCl}^0$ for the solid solute. But NaCl has zero vapour pressure at room temperature — it is non-volatile.

How to avoid:

For a non-volatile solute, the vapour pressure of the solute is zero.
Only the solvent contributes to total vapour pressure.
Total vapour pressure:

P_{total} = x_{solvent} P_{solvent}^0

For NaCl in water:

P_{total} = x_{H_2O} P_{H_2O}^0

Mistake 3: Forgetting that NaCl Dissociates in Water

The error: Students use the formula mass of NaCl to calculate mole fraction, ignoring that NaCl splits into Na⁺ and Cl⁻ ions. This gives an incorrect $x_{H_2O}$ and hence wrong vapour pressure.

How to avoid:

For ionic solutes, use the van’t Hoff factor $i$ $i$ :
- NaCl → $i = 2$ (Na⁺ + Cl⁻)
- Effective moles of particles = $i \times$ moles of NaCl
Then calculate mole fraction of water using total particles (ions + water molecules).

Mistake 4: Assuming Both Components in a Mixture Follow Raoult’s Law When They Don’t

The error: For $CHCl_3$ and $CH_2Cl_2$ , students sometimes assume ideal behaviour without checking. While these two are nearly ideal, many mixtures show positive or negative deviations.

How to avoid:

Ideal solution → $\Delta H_{mix} = 0$ , $\Delta V_{mix} = 0$ …

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Q.Using Raoult's law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.

The core idea

(i) CHCl3(l)CHCl_3(l)CHCl3​(l) and CH2Cl2(l)CH_2Cl_2(l)CH2​Cl2​(l) — two volatile liquids, nearly ideal

(ii) NaCl(s)NaCl(s)NaCl(s) and H2O(l)H_2O(l)H2​O(l) — a non‑volatile solute in a volatile solvent

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(i) $CHCl_3(l)$ and $CH_2Cl_2(l)$ — two volatile liquids, nearly ideal

(ii) $NaCl(s)$ and $H_2O(l)$ — a non‑volatile solute in a volatile solvent

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