Q.Vapour pressure of chloroform ($CHCl_3$) and dichloromethane ($CH_2Cl_2$) at 298 K are 200 mm Hg and 415 mm Hg respectively.

Concept: Raoult's Law and Dalton's Law of Partial Pressures. When two volatile liquids mix, each contributes a partial vapour pressure equal to its mole fraction times its own pure vapour pressure, and the total is their sum.

1. Calculate the moles and liquid phase mole fractions of each component.

   $M_{CHCl_3} = 119.37 \, \text{g/mol}$, $M_{CH_2Cl_2} = 84.93 \, \text{g/mol}$

   $n_{CHCl_3} = \frac{25.5 \, \text{g}}{119.37 \, \text{g/mol}} = 0.2136 \, \text{mol}$

   $n_{CH_2Cl_2} = \frac{40 \, \text{g}}{84.93 \, \text{g/mol}} = 0.4710 \, \text{mol}$

   $X_{CHCl_3} = \frac{0.2136}{0.2136 + 0.4710} = 0.3120$

   $X_{CH_2Cl_2} = \frac{0.4710}{0.2136 + 0.4710} = 0.6880$

2. Calculate the total vapour pressure of the solution using Raoult's Law.

   $P_{total} = P^0_{CHCl_3} X_{CHCl_3} + P^0_{CH_2Cl_2} X_{CH_2Cl_2}$

   $P_{total} = (200 \, \text{mm Hg} \times 0.3120) + (415 \, \text{mm Hg} \times 0.6880)$

   $P_{total} = 62.40 \, \text{mm Hg} + 285.52 \, \text{mm Hg} = 347.92 \, \text{mm Hg} \approx 347.9 \, \text{mm Hg}$ …

When two volatile liquids are mixed, they form a solution, and each component contributes to the total vapour pressure above the solution. For an ideal solution, this contribution is governed by Raoult's Law. An ideal solution is one where the intermolecular forces between like molecules (A-A, B-B) are similar to those between unlike molecules (A-B). Chloroform and dichloromethane are structurally similar and have comparable intermolecular forces (primarily dipole-dipole interactions and London dispersion forces), making their mixture behave very close to an ideal solution.

The core idea is to first determine the composition of the liquid mixture in terms of mole fractions. Then, using Raoult's Law, we calculate the partial vapour pressure exerted by each component. The sum of these partial pressures gives the total vapour pressure of the solution (Dalton's Law). Finally, the mole fraction of a component in the vapour phase is simply its partial pressure divided by the total vapour pressure.

Here's a step-by-step breakdown:

1. Calculate Molar Masses:

   To find the number of moles of each component, we first need their molar masses.

   • For Chloroform ($CHCl_3$): $M_{CHCl_3} = (1 \times 12.01) + (1 \times 1.01) + (3 \times 35.45) = 12.01 + 1.01 + 106.35 = 119.37 \text{ g/mol}$
   • For Dichloromethane ($CH_2Cl_2$): $M_{CH_2Cl_2} = (1 \times 12.01) + (2 \times 1.01) + (2 \times 35.45) = 12.01 + 2.02 + 70.90 = 84.93 \text{ g/mol}$

2. Calculate Moles of Each Component:

   • Moles of $CHCl_3$: $n_{CHCl_3} = \frac{25.5 \text{ g}}{119.37 \text{ g/mol}} \approx 0.2136 \text{ mol}$
   • Moles of $CH_2Cl_2$: $n_{CH_2Cl_2} = \frac{40 \text{ g}}{84.93 \text{ g/mol}} \approx 0.4710 \text{ mol}$

3. Calculate Mole Fractions in the Liquid Phase:

   • Total moles: $n_{total} = 0.2136 + 0.4710 = 0.6846 \text{ mol}$
   • $x_{CHCl_3} = \frac{0.2136}{0.6846} \approx 0.3120$
   • $x_{CH_2Cl_2} = \frac{0.4710}{0.6846} \approx 0.6880$
   Note

   The sum of mole fractions in any phase must always be 1. Here, $0.3120 + 0.6880 = 1.0000$.

4. Calculate Partial Vapour Pressure of Each Component (Raoult's Law):

   $P_i = x_i P_i^0$ …

Method: Raoult's Law + Dalton's Law for an Ideal Binary Liquid Mixture

This problem combines two laws: Raoult's Law (to get each component's partial vapour pressure from the liquid-phase composition) and Dalton's Law of Partial Pressures (to get the vapour-phase composition from the partial pressures).

Steps

Step 1: Convert masses to moles

Step 2: Find liquid-phase mole fractions

Step 3: Apply Raoult's Law to get each partial pressure

Step 4: Add the partial pressures for the total (Dalton's Law)

Step 5: Get vapour-phase mole fractions from Dalton's Law

Final Answer

$\boxed{P_{total} = 347.9 \text{ mm Hg}, \quad y_{CHCl_3} = 0.179, \quad y_{CH_2Cl_2} = 0.821}$ …

Common Mistakes in Raoult's Law Application (CHCl₃ + CH₂Cl₂ Problem)

Here are the most frequent errors students make, with clear explanations of why they happen and how to avoid them.

1. ✗ Forgetting to Convert Mass to Moles First

The Mistake:

Plugging the given masses (25.5 g and 40 g) directly into Raoult's Law without converting to moles.

Why it happens:

Students see "vapour pressure" and "mass" and rush to apply $P = P^\circ \cdot X$ without realising $X$ is mole fraction, not mass fraction.

How to avoid:

Always write the mole calculation step explicitly:

• Molar mass of CHCl₃ = $12 + 1 + 3 \times 35.5 = 119.5 \, \text{g/mol}$

  Moles of CHCl₃ = $\frac{25.5}{119.5} = 0.2134 \, \text{mol}$

• Molar mass of CH₂Cl₂ = $12 + 2 + 2 \times 35.5 = 85 \, \text{g/mol}$

  Moles of CH₂Cl₂ = $\frac{40}{85} = 0.4706 \, \text{mol}$

Rule: Mass → Moles → Mole fraction → Raoult's Law. Never skip step 1.

2. ✗ Using Wrong Mole Fraction Formula

The Mistake:

Using $X_A = \frac{n_A}{n_A}$ (always 1) or $X_A = \frac{n_A}{n_A + n_B}$ incorrectly (e.g., swapping numerator/denominator).

Why it happens:

Confusion between "mole fraction of A in solution" vs "mole fraction of A in vapour".

How to avoid:

• In solution: $X_A = \frac{n_A}{n_A + n_B}$
• In vapour: $Y_A = \frac{P_A}{P_{\text{total}}}$ (from Dalton's Law)

For this problem:

$X_{\text{CHCl}_3} = \frac{0.2134}{0.2134 + 0.4706} = 0.312$

$X_{\text{CH}_2\text{Cl}_2} = 1 - 0.312 = 0.688$

Check: Sum of mole fractions in solution = 1. Always verify.

3. ✗ Confusing Partial Pressure with Vapour Pressure of Pure Component

The Mistake:

Writing $P_{\text{CHCl}_3} = 200 \, \text{mm Hg}$ (the pure vapour pressure) instead of $P_{\text{CHCl}_3} = X_{\text{CHCl}_3} \times 200$.

Why it happens:

Students memorise Raoult's Law as $P = P^\circ$ instead of $P = X \cdot P^\circ$.

How to avoid:

Write the full form every time:

$P_A = X_A \cdot P_A^\circ$

For this problem:

$P_{\text{CHCl}_3} = 0.312 \times 200 = 62.4 \, \text{mm Hg}$

$P_{\text{CH}_2\text{Cl}_2} = 0.688 \times 415 = 285.5 \, \text{mm Hg}$

Mnemonic: "Partial pressure = fraction of molecules × pressure if pure"

4. ✗ Forgetting to Add Partial Pressures for Total Vapour Pressure

The Mistake:

Stopping after calculating one partial pressure, or subtracting instead of adding.

Why it happens:

Students think "vapour pressure of solution" means just one component's contribution.

How to avoid:

Raoult's Law for a binary solution:

$P_{\text{total}} = P_A + P_B = X_A P_A^\circ + X_B P_B^\circ$

Here:

$P_{\text{total}} = 62.4 + 285.5 = 347.9 \, \text{mm Hg}$

Check: Total pressure must lie between the two pure vapour pressures (200 and 415). 347.9 is between — correct!

5. ✗ Using Solution Mole Fractions for Vapour Phase

The Mistake:

Saying mole fraction of CHCl₃ in vapour = 0.312 (the solution value).

Why it happens:

Not distinguishing between liquid phase composition and vapour phase composition.

How to avoid:

Vapour phase mole fractions come from Dalton's Law:

$Y_A = \frac{P_A}{P_{\text{total}}}$

For this problem:

$Y_{\text{CHCl}_3} = \frac{62.4}{347.9} = 0.179$ …