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

Options

A. 3mol
B. 2mol
C. 1mol
D. 0.5mol

Explanation

To solve the problem of how many moles of \( K_2SO_4 \) 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
- Lowering of vapor pressure, \( \Delta P = 10 \, \text{mm Hg} \)
- Vapor pressure of pure water, \( P_0 = 50 \, \text{mm Hg} \)
- Moles of solvent (water), \( n_{solvent} = 12 \, \text{moles} \)

### Step 2: Use Raoult's Law
According to Raoult's Law, the relative lowering of vapor pressure is given by:
\[ \frac{\Delta P}{P_0} = i \cdot X_{solute} \]
Where:
- \( i \) is the van 't Hoff factor (number of particles the solute dissociates into)
- \( X_{solute} \) is the mole fraction of the solute

### Step 3: Calculate the mole fraction of the solute
The mole fraction of the solute can be expressed as:
\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]
Where \( n_{solute} \) is the number of moles of \( K_2SO_4 \) we need to find.

### Step 4: Determine the van 't Hoff factor \( i \)
For \( K_2SO_4 \), it dissociates into:
\[ K_2SO_4 \rightarrow 2K^+ + SO_4^{2-} \]
Thus, \( i = 3 \) (2 potassium ions and 1 sulfate ion).

### Step 5: Substitute values into Raoult's Law
Substituting the known values into Raoult's Law:
\[ \frac{10}{50} = 3 \cdot \frac{n_{solute}}{n_{solute} + 12} \]
This simplifies to:
\[ \frac{1}{5} = 3 \cdot \frac{n_{solute}}{n_{solute} + 12} \]

### Step 6: Cross-multiply to solve for \( n_{solute} \)
Cross-multiplying gives:
\[ n_{solute} + 12 = 15 \cdot n_{solute} \]
Rearranging gives:
\[ 12 = 15n_{solute} - n_{solute} \]
\[ 12 = 14n_{solute} \]
\[ n_{solute} = \frac{12}{14} = \frac{6}{7} \approx 0.857 \, \text{moles} \]

### Step 7: Final calculation
To find the number of moles of \( K_2SO_4 \):
\[ n_{solute} \approx 0.857 \, \text{moles} \]

### Conclusion
Thus, the number of moles of \( K_2SO_4 \) to be dissolved in 12 moles of water to lower its vapor pressure by 10 mm Hg is approximately **0.857 moles**.