At 300 K , when a solute is aded to a solvent ,its vapour pressure over mercury reduces from 50 mm to 45 mm . The value of mole fraction of solute wil be _____.

Options

A. 0.005
B. `0.010`
C. `0.100`
D. `0.900`

Answer

To find the mole fraction of the solute when a solute is added to a solvent, we can use the formula for relative lowering of vapor pressure. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Vapor pressure of pure solvent (P₀) = 50 mm
- Vapor pressure of solution (P) = 45 mm

### Step 2: Calculate the lowering of vapor pressure
The lowering of vapor pressure can be calculated as:
\[ \Delta P = P₀ - P = 50 \, \text{mm} - 45 \, \text{mm} = 5 \, \text{mm} \]

### Step 3: Use the formula for relative lowering of vapor pressure
The formula for relative lowering of vapor pressure is given by:
\[ \frac{\Delta P}{P₀} = \text{mole fraction of solute (X₁)} \]
Substituting the values we have:
\[ \frac{5 \, \text{mm}}{50 \, \text{mm}} = \text{mole fraction of solute (X₁)} \]

### Step 4: Calculate the mole fraction of solute
Now, calculate the mole fraction:
\[ \text{mole fraction of solute (X₁)} = \frac{5}{50} = 0.1 \]

### Step 5: Conclusion
Thus, the mole fraction of the solute is:
\[ \text{Mole fraction of solute} = 0.1 \]

### Final Answer
The value of the mole fraction of solute will be **0.1**.

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