Q.100 g of liquid A (molar mass 140 g mol$^{-1}$) was dissolved in 1000 g of liquid B (molar mass 180 g mol$^{-1}$). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Concept: Raoult’s Law for a Binary Solution of Two Volatile Liquids.

Step 1: Find moles of each component

Moles of A:

$n_A = \frac{100}{140} = 0.7143$ mol

Moles of B:

$n_B = \frac{1000}{180} = 5.5556$ mol

Step 2: Mole fractions in the liquid phase

$x_A = \frac{0.7143}{0.7143 + 5.5556} = \frac{0.7143}{6.2699} = 0.1139$

$x_B = 1 - x_A = 0.8861$

Step 3: Apply Raoult’s law for total vapour pressure

$P_{\text{total}} = P_A^\circ x_A + P_B^\circ x_B$

Given $P_{\text{total}} = 475$ torr and $P_B^\circ = 500$ torr:

$475 = P_A^\circ (0.1139) + 500(0.8861)$

$475 = 0.1139 P_A^\circ + 443.05$

$0.1139 P_A^\circ = 31.95$ …

1. Moles.

2. Mole fractions.

3. Total pressure (Raoult's law).

$475 = \frac{9}{79} P_A^\circ + \frac{70}{79}(500) = \frac{9}{79} P_A^\circ + 443.04$ …

Method: Raoult's Law for an Ideal Solution of Two Volatile Liquids

Concept (Why this works)

Both A and B are volatile, so both contribute to the total vapour pressure. By Raoult's law, each component's partial pressure is its own mole fraction times its own pure vapour pressure, and by Dalton's law the total is their sum:

Here $P_B^\circ$ (pure B) and $P_{\text{total}}$ (the solution) are given, so this single equation can be solved for the one unknown, $P_A^\circ$.

Steps

Step 1: Find moles of each liquid

Step 2: Find mole fractions

Step 3: Apply Raoult's law to the total pressure and solve for $P_A^\circ$

$475 = 0.114\,P_A^\circ + 443.0$ …

Here are the common mistakes students make on this exact type of problem (finding the vapour pressure of pure A from the total vapour pressure of an ideal binary solution), and how to avoid each.

1. Forgetting to Convert Mass to Moles First

The Mistake: Using the given masses (100 g of A, 1000 g of B) directly instead of moles.

How to avoid:

• $n_A = \dfrac{100}{140} \approx 0.714$ mol
• $n_B = \dfrac{1000}{180} \approx 5.556$ mol

2. Not Realising There Are Two Unknowns and Only One Given Total Pressure

The Mistake: Trying to find $P_A^\circ$ without setting up Raoult's law properly, or assuming $P_A^\circ$ is somehow already known.

How to avoid: Recognise that $P_B^\circ$ (500 torr) is given directly, but $P_A^\circ$ is unknown -- it must be found from the one piece of information that links it to something known: the total pressure of the solution, $P_{\text{total}} = 475$ torr.

This is a single linear equation in the single unknown $P_A^\circ$, solvable directly.

3. Swapping Which Mole Fraction Goes With Which Vapour Pressure

The Mistake: Writing $P_{\text{total}} = x_B P_A^\circ + x_A P_B^\circ$ (mole fractions swapped).

How to avoid: Always match a component's own mole fraction to its own pure vapour pressure: $x_A$ pairs with $P_A^\circ$, $x_B$ pairs with $P_B^\circ$.

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