The vapour pressure of pure liquid solvent A is `0.80`atm.When a non volatile substances B is added to the solvent, its vapour pressure drops to `0.60`atm. Mole fraction of the components B in the solution is:

Options

A. `0.50`
B. `0.25`
C. `0.75`
D. `0.40`

Explanation

To find the mole fraction of the non-volatile substance B in the solution, we can use the concept of relative lowering of vapor pressure. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Vapor pressure of pure solvent A (P₀) = 0.80 atm
- Vapor pressure of the solution (P) = 0.60 atm

### Step 2: Calculate the change in vapor pressure (ΔP)
ΔP = P₀ - P
ΔP = 0.80 atm - 0.60 atm
ΔP = 0.20 atm

### Step 3: Use the formula for relative lowering of vapor pressure
The relative lowering of vapor pressure is given by the formula:
\[ \frac{\Delta P}{P_0} = \frac{n_B}{n_A + n_B} \]
Where:
- \(n_B\) = number of moles of solute B
- \(n_A\) = number of moles of solvent A

### Step 4: Express the mole fraction of B
The mole fraction of solute B (X_B) can be expressed as:
\[ X_B = \frac{n_B}{n_A + n_B} \]

### Step 5: Substitute ΔP and P₀ into the relative lowering formula
Substituting the values we have:
\[ \frac{0.20}{0.80} = \frac{n_B}{n_A + n_B} \]
This simplifies to:
\[ 0.25 = \frac{n_B}{n_A + n_B} \]

### Step 6: Rearranging the equation
From the equation above, we can rearrange it to find \(n_B\):
\[ 0.25(n_A + n_B) = n_B \]
\[ 0.25n_A + 0.25n_B = n_B \]
\[ 0.25n_A = n_B - 0.25n_B \]
\[ 0.25n_A = 0.75n_B \]
\[ n_B = \frac{0.25}{0.75}n_A \]
\[ n_B = \frac{1}{3}n_A \]

### Step 7: Find the mole fraction of B
Now substituting \(n_B\) back into the mole fraction formula:
\[ X_B = \frac{n_B}{n_A + n_B} = \frac{\frac{1}{3}n_A}{n_A + \frac{1}{3}n_A} \]
\[ X_B = \frac{\frac{1}{3}n_A}{\frac{4}{3}n_A} = \frac{1}{4} \]

### Final Answer
The mole fraction of component B in the solution is \(X_B = \frac{1}{4}\) or 0.25.

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