What will be the number of moles of `K_(2)SO_(4)` to be dissolved in 12 mol of water to lower . Its vapour pressure by 10 mm Hg at a temperature at which vapours pressure of pure water is 50 mm .

Answer

To solve the problem of determining the number of moles of potassium sulfate (K₂SO₄) needed to lower the vapor pressure of water by 10 mm Hg, we can use the concept of colligative properties, specifically the relative lowering of vapor pressure. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- Vapor pressure of pure water (P₀) = 50 mm Hg
- Lowering of vapor pressure (P₀ - P) = 10 mm Hg
- Moles of water (solvent) = 12 mol

### Step 2: Calculate the Vapor Pressure of the Solution
Using the information provided:
\[ P = P₀ - (P₀ - P) = 50 \, \text{mm Hg} - 10 \, \text{mm Hg} = 40 \, \text{mm Hg} \]

### Step 3: Apply the Formula for Relative Lowering of Vapor Pressure
The formula for relative lowering of vapor pressure is:
\[ \frac{P₀ - P}{P₀} = \text{mole fraction of solute} \]

Substituting the values we have:
\[ \frac{10 \, \text{mm Hg}}{50 \, \text{mm Hg}} = \frac{1}{5} \]

### Step 4: Relate Mole Fraction to Moles of Solute
Let the number of moles of K₂SO₄ be \( X \). The mole fraction of the solute (K₂SO₄) can be expressed as:
\[ \text{Mole fraction of solute} = \frac{X}{X + 12} \]

Setting this equal to the relative lowering we calculated:
\[ \frac{X}{X + 12} = \frac{1}{5} \]

### Step 5: Cross-Multiply to Solve for X
Cross-multiplying gives:
\[ 5X = X + 12 \]

### Step 6: Simplify the Equation
Rearranging the equation:
\[ 5X - X = 12 \]
\[ 4X = 12 \]

### Step 7: Solve for X
Dividing both sides by 4:
\[ X = \frac{12}{4} = 3 \]

### Conclusion
The number of moles of potassium sulfate (K₂SO₄) needed to be dissolved in 12 moles of water to lower its vapor pressure by 10 mm Hg is **3 moles**.

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