Exhaustive Guide: Relative Lowering of Vapour Pressure | Chemca

Exhaustive Guide: Relative Lowering of Vapour Pressure

The fundamental Colligative Property. Master Raoult's Law, the Ostwald-Walker dynamic method, and complex physical chemistry numericals for JEE and NEET.

1. Introduction: The Concept of Vapour Pressure

Welcome back to Chemca.in. To master the colligative properties of solutions—such as boiling point elevation, freezing point depression, and osmotic pressure—we must first conquer their thermodynamic progenitor: the Relative Lowering of Vapour Pressure (RLVP).

In a closed vessel, the molecules of a liquid possess varying kinetic energies. Those at the surface with sufficient energy overcome intermolecular forces and escape into the gas phase (evaporation). Simultaneously, gaseous molecules collide with the surface and re-enter the liquid phase (condensation). When the rate of evaporation precisely equals the rate of condensation, a state of dynamic equilibrium is reached. The pressure exerted by these vapors on the liquid surface at this equilibrium is called the Vapour Pressure of the liquid at that specific temperature.

What happens when we take a pure, volatile solvent (like water or benzene) and dissolve a non-volatile solute (like glucose, urea, or $NaCl$) into it? The vapour pressure of the resulting solution plummets. Let us explore the molecular and thermodynamic reasons behind this fascinating phenomenon.

2. The Mechanism: Why Does Vapour Pressure Drop?

The lowering of vapour pressure is fundamentally a surface phenomenon, brilliantly explained by the Kinetic Molecular Theory and Raoult's Law.

Figure 1: In a pure solvent, the entire surface is occupied by volatile molecules. In a solution, non-volatile solute particles occupy a fraction of the surface, physically blocking solvent molecules from escaping, thus lowering the vapour pressure.

2.1. The Kinetic Molecular Explanation

Evaporation happens exclusively at the surface. In a pure solvent, $100\%$ of the surface area is occupied by solvent molecules capable of vaporizing. When a non-volatile solute is dissolved, its particles are uniformly distributed throughout the bulk and the surface. These non-volatile particles act as physical roadblocks. The fraction of the surface occupied by solvent molecules decreases, the rate of evaporation decreases, and consequently, the equilibrium vapour pressure drops.

2.2. The Thermodynamic Explanation (Entropy)

A pure liquid is a highly ordered state compared to its vapor. The process of vaporization increases entropy ($\Delta S > 0$). However, when a solute is dissolved into the liquid, the resulting solution is already more disordered (higher entropy) than the pure liquid. Therefore, the drive (the entropy gain) to transition from the liquid phase to the gas phase is lessened. This thermodynamic reluctance manifests as a lower vapour pressure.

3. Mathematical Derivation of RLVP (Raoult's Law)

In 1886, the French chemist François-Marie Raoult established the quantitative relationship between the vapour pressure of a solution and its concentration. Raoult's Law states that the partial vapour pressure of any volatile component in a solution is directly proportional to its mole fraction in the solution.

Let us consider a binary solution containing a volatile solvent (Component 1) and a non-volatile solute (Component 2).

Let ${P_1^{\circ}}$ be the vapour pressure of the pure solvent.
Let ${P_s}$ (or $P_1$) be the vapour pressure of the solution.
Let $x_1$ and $x_2$ be the mole fractions of the solvent and solute, respectively.

Since the solute is non-volatile, it exerts no vapour pressure. The entire vapour pressure of the solution is due only to the solvent:

$$ {P_s} = {P_1^{\circ}} \times x_1 $$

We know that the sum of mole fractions in a binary mixture is exactly 1 ($x_1 + x_2 = 1$). Substituting $x_1 = 1 - x_2$ into the equation:

$$ {P_s} = {P_1^{\circ}} (1 - x_2) $$ $$ {P_s} = {P_1^{\circ}} - {P_1^{\circ}} x_2 $$

Rearranging the terms, we get the expression for the Lowering of Vapour Pressure:

$$ {P_1^{\circ}} - {P_s} = {P_1^{\circ}} x_2 $$

To find the Relative Lowering of Vapour Pressure (RLVP), we divide both sides by the original vapour pressure of the pure solvent (${P_1^{\circ}}$):

                    The Fundamental Equation of RLVP:
                    $$ \frac{{P_1^{\circ}} - {P_s}}{{P_1^{\circ}}} = x_2 $$
                

This beautiful equation tells us that the relative lowering of vapour pressure is a colligative property—it is equal to the mole fraction of the solute ($x_2$), completely independent of what the solute actually is!

4. Determination of Unknown Molar Mass

The primary laboratory application of RLVP is determining the molecular weight of unknown non-volatile substances (like polymers or newly synthesized drugs). We can expand the mole fraction term ($x_2$) in terms of masses and molar masses.

Let $w_2$ and $M_2$ be the mass and molar mass of the solute. Let $w_1$ and $M_1$ be the mass and molar mass of the solvent. The number of moles are $n_2 = w_2/M_2$ and $n_1 = w_1/M_1$.

$$ x_2 = \frac{n_2}{n_1 + n_2} $$

For Dilute Solutions: In a very dilute solution, the number of moles of solute ($n_2$) is negligible compared to the moles of solvent ($n_1$). Thus, the denominator $(n_1 + n_2) \approx n_1$.

$$ \frac{{P_1^{\circ}} - {P_s}}{{P_1^{\circ}}} \approx \frac{n_2}{n_1} = \frac{(w_2 / M_2)}{(w_1 / M_1)} $$

                    Working Formula for Molar Mass (Dilute Solutions):
                    $$ \frac{\Delta P}{{P_1^{\circ}}} = \frac{w_2 \times M_1}{M_2 \times w_1} $$
                

Important Note for JEE/NEET: If the solution is concentrated, making the dilute approximation leads to mathematical errors. A brilliant algebraic trick used in competitive exams is to relate the lowering directly to the solution's vapour pressure (${P_s}$) instead of the pure solvent's (${P_1^{\circ}}$). Without approximation, the exact formula is:

                    Exact Formula (Valid for all concentrations):
                    $$ \frac{{P_1^{\circ}} - {P_s}}{{P_s}} = \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1} $$
                

5. The Ostwald-Walker Dynamic Method

Measuring vapour pressure directly with a barometer is prone to massive errors from dissolved air or temperature fluctuations. To measure RLVP with extreme precision, physical chemists use the Ostwald-Walker Method.

The Setup: A stream of perfectly dry air is passed through two sets of bulbs connected in series. The first set contains the Solution, and the second set contains the Pure Solvent. Finally, the air passes through a $U$-tube packed with anhydrous Calcium Chloride ($CaCl_2$), a powerful desiccant.

Set 1 (Solution Bulbs): As dry air passes through the solution, it absorbs solvent vapour until it is saturated up to the vapour pressure of the solution (${P_s}$). The solution bulbs lose mass proportional to ${P_s}$.
Loss in mass of Solution $\propto {P_s}$
Set 2 (Solvent Bulbs): The air enters the pure solvent already partially saturated (at ${P_s}$). Since the pure solvent has a higher vapour pressure (${P_1^{\circ}}$), the air absorbs *more* vapour to reach full saturation at ${P_1^{\circ}}$. The solvent bulbs lose mass proportional to the difference.
Loss in mass of Solvent $\propto ({P_1^{\circ}} - {P_s})$
U-Tube (Desiccant): The $CaCl_2$ tube absorbs all the moisture from the air. Its mass increases proportional to the total vapour pressure (${P_1^{\circ}}$).
Gain in mass of $CaCl_2 \propto {P_1^{\circ}}$

From these proportionalities, we can calculate the RLVP directly without ever using a pressure gauge:

                    Ostwald-Walker Formula:
                    $$ \frac{{P_1^{\circ}} - {P_s}}{{P_1^{\circ}}} = \frac{\text{Loss in mass of Solvent bulbs}}{\text{Total gain in mass of U-tube}} $$
                

6. Modification for Electrolytes: The Van't Hoff Factor ($i$)

If the non-volatile solute is an electrolyte (like $NaCl$, $K_2SO_4$, or $MgCl_2$), it will dissociate into multiple ions in the solvent. Since RLVP is a colligative property, it depends on the actual number of particles in the solution, not just the moles added. We must modify the Raoult's law equation using the Van't Hoff factor ($i$).

$$ \frac{{P_1^{\circ}} - {P_s}}{{P_1^{\circ}}} = i \times x_2 \quad (\text{For dilute solutions}) $$

For complete dissociation ($\alpha = 1$), $i$ equals the number of ions produced (e.g., $i = 3$ for $CaCl_2$). For partial dissociation, $i = 1 + \alpha(n - 1)$. If this factor is ignored in numerical problems involving salts, the calculated molar mass will be completely wrong (abnormal molar mass).

7. Masterclass: Solved Numericals (CBSE to JEE Advanced)

Problem 1: Standard Molar Mass Determination (CBSE)

Question: The vapour pressure of pure water at $298 \text{ K}$ is $23.8 \text{ mmHg}$. $50 \text{ g}$ of urea ($NH_2CONH_2$) is dissolved in $850 \text{ g}$ of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Strategy: Urea is a non-electrolyte ($i=1$). Find the moles of water and urea, calculate the mole fraction of the solute, and apply Raoult's Law.

Step 1: Calculate moles
Molar mass of Urea ($M_2$) = $60 \text{ g/mol}$. Moles of urea ($n_2$) = $50 / 60 = 0.833 \text{ mol}$.
Molar mass of Water ($M_1$) = $18 \text{ g/mol}$. Moles of water ($n_1$) = $850 / 18 = 47.22 \text{ mol}$.

Step 2: Calculate Mole Fraction ($x_2$)
$x_2 = \frac{n_2}{n_1 + n_2} = \frac{0.833}{47.22 + 0.833} = \frac{0.833}{48.053} = 0.0173$

Step 3: Calculate RLVP and ${P_s}$
RLVP = $x_2 = 0.0173$.
$\frac{23.8 - {P_s}}{23.8} = 0.0173 \implies 23.8 - {P_s} = 0.411 \implies {P_s} = 23.389 \text{ mmHg}$.

Final Answer: RLVP = $0.0173$, Vapour Pressure of solution = $23.39 \text{ mmHg}$.

Problem 2: The Exact Formula Application (JEE Main)

Question: The vapour pressure of pure liquid A is $40 \text{ torr}$ at $310 \text{ K}$. The vapour pressure of this liquid in a solution with a non-volatile liquid B is $32 \text{ torr}$. Calculate the mole fraction of B in the solution.

Strategy: Direct application of the RLVP equation. No mass conversions are needed.

Step 1: Identify given values
${P_1^{\circ}} = 40 \text{ torr}$
${P_s} = 32 \text{ torr}$

Step 2: Apply RLVP Formula
$\frac{{P_1^{\circ}} - {P_s}}{{P_1^{\circ}}} = x_2$
$\frac{40 - 32}{40} = x_2$
$\frac{8}{40} = x_2 \implies x_2 = 0.2$

Final Answer: Mole fraction of B ($x_2$) = $0.2$.

Problem 3: Van't Hoff Factor Integration (NEET)

Question: A solution containing $30 \text{ g}$ of a non-volatile, strong electrolyte $AB_2$ in $90 \text{ g}$ of water has a vapour pressure of $2.8 \text{ kPa}$ at $298 \text{ K}$. If the vapour pressure of pure water at this temperature is $3.0 \text{ kPa}$, calculate the molar mass of the electrolyte $AB_2$. (Assume $100\%$ dissociation).

Strategy: Since $AB_2$ is a strong electrolyte, it dissociates completely into $A^{2+}$ and $2B^-$. Therefore, the Van't Hoff factor $i = 3$. We must use the exact formula to avoid dilution approximation errors.

Step 1: Determine 'i' and moles of solvent
$i = 3$.
Moles of water ($n_1$) = $90 / 18 = 5 \text{ mol}$.
Let molar mass of $AB_2$ be $M_2$. Moles of solute ($n_2$) = $30 / M_2$.

Step 2: Apply Exact Formula with $i$
$\frac{{P_1^{\circ}} - {P_s}}{{P_s}} = \frac{i \times n_2}{n_1}$
$\frac{3.0 - 2.8}{2.8} = \frac{3 \times (30 / M_2)}{5}$
$\frac{0.2}{2.8} = \frac{90}{5 \times M_2}$
$\frac{1}{14} = \frac{18}{M_2}$

Step 3: Solve for $M_2$
$M_2 = 14 \times 18 = 252 \text{ g/mol}$.

Final Answer: Molar mass of $AB_2$ = $252 \text{ g/mol}$.

Problem 4: Ostwald-Walker Method (JEE Advanced)

Question: In an Ostwald-Walker experiment, a stream of dry air is passed through a solution of $6.5 \text{ g}$ of a non-volatile solute in $100 \text{ g}$ of water, then through pure water, and finally through a $U$-tube containing anhydrous $CaCl_2$. The loss in mass of the solution bulbs is $2.0 \text{ g}$ and the loss in mass of the pure water bulbs is $0.05 \text{ g}$. Calculate the molar mass of the solute.

Strategy: Use the proportionalities of the Ostwald-Walker method. Loss in solution $\propto {P_s}$. Loss in solvent $\propto ({P_1^{\circ}} - {P_s})$. Total gain in U-tube = Sum of losses $\propto {P_1^{\circ}}$. Then use the RLVP formula.

Step 1: Determine Mass Changes
Loss in solution bulbs = $2.0 \text{ g}$ ($\propto {P_s}$)
Loss in solvent bulbs = $0.05 \text{ g}$ ($\propto {P_1^{\circ}} - {P_s}$)
Total mass absorbed by $CaCl_2$ = $2.0 + 0.05 = 2.05 \text{ g}$ ($\propto {P_1^{\circ}}$).

Step 2: Calculate RLVP
$\frac{{P_1^{\circ}} - {P_s}}{{P_1^{\circ}}} = \frac{\text{Loss in solvent bulbs}}{\text{Total gain in U-tube}}$
$\frac{\Delta P}{{P_1^{\circ}}} = \frac{0.05}{2.05} \approx 0.02439$.

Step 3: Calculate Molar Mass
Using the dilute formula: $\frac{\Delta P}{{P_1^{\circ}}} = \frac{w_2 \times M_1}{M_2 \times w_1}$
$0.02439 = \frac{6.5 \times 18}{M_2 \times 100}$
$M_2 = \frac{117}{0.02439 \times 100} = \frac{117}{2.439} \approx 47.97 \text{ g/mol}$.

Final Answer: Molar mass of the solute $\approx 48 \text{ g/mol}$.

8. Conclusion

The Relative Lowering of Vapour Pressure is the thermodynamic grandparent of all colligative properties. Because adding a non-volatile solute physically and entropically suppresses the escaping tendency of the solvent molecules, the vapor pressure inevitably drops. This single principle cascaded into the realization that such solutions must be boiled at higher temperatures (Elevation of Boiling Point) and cooled to lower temperatures (Depression in Freezing Point) to reach equilibrium.

For JEE and NEET aspirants, the two most critical areas to practice are mastering the Exact Formula $\frac{{P^{\circ}} - {P_s}}{{P_s}} = \frac{n_2}{n_1}$ to avoid dilution approximation errors, and remembering to implement the Van't Hoff factor ($i$) whenever an ionic salt is mentioned. Keep this guide bookmarked for your final electro-physical chemistry revisions!

9. Frequently Asked Questions (FAQs)

Q1. Why does Raoult's Law only apply to non-volatile solutes in this context?

If the solute is volatile (like alcohol in water), it will also evaporate and exert its own vapour pressure. In that scenario, the total vapour pressure of the solution is the sum of the partial pressures of both components ($P_{\text{total}} = P_A + P_B$). The specific formula for RLVP ($\Delta P / {P_1^{\circ}} = x_2$) assumes the solute's vapor pressure is strictly zero.

Q2. Is the Lowering of Vapour Pressure a colligative property?

No! The absolute lowering ($\Delta P = {P_1^{\circ}} - {P_s}$) is not a true colligative property because its value changes depending on the identity and temperature of the pure solvent. Only the Relative Lowering ($\Delta P / {P_1^{\circ}}$) is a true colligative property because it equals the mole fraction of the solute, completely independent of the solvent's specific vapor pressure curve.

Q3. Why is the Ostwald-Walker method preferred over a barometer?

Barometers measure total pressure. In a laboratory setting, the total pressure above a liquid includes the vapor pressure of the liquid *plus* any dissolved atmospheric gases (air) that escape. The Ostwald-Walker dynamic method measures mass loss due strictly to the evaporation of the solvent, completely eliminating errors caused by dissolved air.

Q4. How does temperature affect the Relative Lowering of Vapour Pressure?

Surprisingly, temperature has virtually no effect on the relative lowering of vapour pressure! While raising the temperature exponentially increases the pure vapour pressure (${P_1^{\circ}}$) and the solution's vapour pressure (${P_s}$), their relative ratio ($\Delta P / {P_1^{\circ}}$) remains constant because it is strictly equal to the mole fraction ($x_2$), which is independent of temperature.

Q5. When should I use the 'Exact Formula' instead of the standard dilute formula?

The standard formula ($\Delta P / {P^{\circ}} \approx n_2 / n_1$) assumes $n_2$ is negligible compared to $n_1$. You should use this ONLY if the solute concentration is explicitly stated as very dilute (usually $< 5\%$ by mass). For concentrated solutions, or in strict JEE Advanced numericals where precision matters, always use the exact formula ($\Delta P / {P_s} = n_2 / n_1$) derived algebraically without approximations.

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