An aqueous solution contianing `20%` by weight of liquid X (mol.wt. `=140`) has a vapour pressure `160mm` at `60^(@)C`. Calculate the vapour pressure of pure liquid 'X' if the vapour pressure of water is 150mm at `60^(@)C`.

Answer

To solve the problem step by step, we will use Raoult's Law and the concept of mole fractions.

### Step 1: Determine the mass of the components in the solution
Given that the solution is 20% by weight of liquid X, we can assume we have 100 grams of the solution. Therefore:
- Mass of liquid X = 20 grams
- Mass of water = 100 grams - 20 grams = 80 grams

### Step 2: Calculate the number of moles of each component
Using the molecular weights:
- Molecular weight of liquid X = 140 g/mol
- Molecular weight of water = 18 g/mol

Now, we can calculate the number of moles:
- Moles of liquid X = mass / molar mass = 20 g / 140 g/mol = 0.1429 moles
- Moles of water = mass / molar mass = 80 g / 18 g/mol = 4.4444 moles

### Step 3: Calculate the mole fractions
The total number of moles in the solution is:
Total moles = Moles of liquid X + Moles of water = 0.1429 + 4.4444 = 4.5873 moles

Now, we can calculate the mole fractions:
- Mole fraction of liquid X (solute) = Moles of liquid X / Total moles = 0.1429 / 4.5873 = 0.0311
- Mole fraction of water (solvent) = Moles of water / Total moles = 4.4444 / 4.5873 = 0.9689

### Step 4: Apply Raoult's Law
According to Raoult's Law:
\[ P_{solution} = (X_{solvent} \cdot P^0_{solvent}) + (X_{solute} \cdot P^0_{solute}) \]

Where:
- \( P_{solution} \) = vapor pressure of the solution = 160 mm
- \( X_{solvent} \) = mole fraction of water = 0.9689
- \( P^0_{solvent} \) = vapor pressure of pure water at 60°C = 150 mm
- \( X_{solute} \) = mole fraction of liquid X = 0.0311
- \( P^0_{solute} \) = vapor pressure of pure liquid X (which we need to find)

Substituting the values into Raoult's Law:
\[ 160 = (0.9689 \cdot 150) + (0.0311 \cdot P^0_{solute}) \]

### Step 5: Solve for \( P^0_{solute} \)
Calculating the first term:
\[ 0.9689 \cdot 150 = 145.335 \]

Now substituting back into the equation:
\[ 160 = 145.335 + (0.0311 \cdot P^0_{solute}) \]
\[ 160 - 145.335 = 0.0311 \cdot P^0_{solute} \]
\[ 14.665 = 0.0311 \cdot P^0_{solute} \]

Now, divide both sides by 0.0311 to find \( P^0_{solute} \):
\[ P^0_{solute} = \frac{14.665}{0.0311} \approx 471.5 \text{ mm} \]

### Final Answer
The vapor pressure of pure liquid X is approximately **471.5 mm**.

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