Moles of` Na_(2)SO_(4)` to be dissoved in 12 mole water to lower its vapour pressure by 10 mm Hg at a temperature at which vapour pressure of pure water is 50 mm is:

Options

A. 1.5 mole
B. 2 mole
C. 1 mole
D. 3 mole

Answer

To solve the problem of how many moles of Na₂SO₄ need to be dissolved in 12 moles of water to lower its vapor pressure by 10 mm Hg, we can follow these steps:

### Step 1: Understand the given data
- Vapor pressure of pure water (P₀) = 50 mm Hg
- Decrease in vapor pressure (ΔP) = 10 mm Hg
- Moles of water (n₁) = 12 moles

### Step 2: Calculate the vapor pressure of the solution
The vapor pressure of the solution (P) can be calculated using the formula:
\[ P = P₀ - ΔP \]
Substituting the values:
\[ P = 50 \, \text{mm Hg} - 10 \, \text{mm Hg} = 40 \, \text{mm Hg} \]

### Step 3: Use Raoult's Law
According to Raoult's Law, the vapor pressure of the solvent in the solution is given by:
\[ P = x_{H₂O} \cdot P₀ \]
Where \( x_{H₂O} \) is the mole fraction of water. Rearranging gives:
\[ x_{H₂O} = \frac{P}{P₀} \]
Substituting the values:
\[ x_{H₂O} = \frac{40 \, \text{mm Hg}}{50 \, \text{mm Hg}} = \frac{4}{5} \]

### Step 4: Relate mole fraction to moles
The mole fraction of water can also be expressed as:
\[ x_{H₂O} = \frac{n_{H₂O}}{n_{H₂O} + n_{Na₂SO₄}} \]
Where \( n_{H₂O} \) is the number of moles of water (12 moles) and \( n_{Na₂SO₄} \) is the number of moles of Na₂SO₄ we need to find.

### Step 5: Set up the equation
Substituting the known values into the mole fraction equation:
\[ \frac{12}{12 + n_{Na₂SO₄}} = \frac{4}{5} \]

### Step 6: Cross-multiply to solve for \( n_{Na₂SO₄} \)
Cross-multiplying gives:
\[ 5 \cdot 12 = 4 \cdot (12 + n_{Na₂SO₄}) \]
\[ 60 = 48 + 4n_{Na₂SO₄} \]

### Step 7: Isolate \( n_{Na₂SO₄} \)
Rearranging the equation:
\[ 60 - 48 = 4n_{Na₂SO₄} \]
\[ 12 = 4n_{Na₂SO₄} \]
\[ n_{Na₂SO₄} = \frac{12}{4} = 3 \]

### Conclusion
Thus, the number of moles of Na₂SO₄ that need to be dissolved is **3 moles**.

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