`6g` of urea ( molecular weight `=60)` was dissolved in `9.9` moles of water. If the vapour pressure of pure water is `P^(@)`, the vapour pressure of solution is `:`

Options

A. `0.10 P^(@)`
B. `1.10P^(@)`
C. `0.90P^(@)`
D. `0.99P^(@)`

Explanation

To solve the problem of finding the vapor pressure of a solution of urea in water, we will follow these steps:

### Step 1: Calculate the moles of urea
Given:
- Mass of urea = 6 g
- Molecular weight of urea = 60 g/mol

Using the formula for moles:
\[ \text{Moles of urea} (n) = \frac{\text{mass}}{\text{molecular weight}} = \frac{6 \, \text{g}}{60 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 2: Identify the moles of solvent (water)
Given:
- Moles of water = 9.9 mol

### Step 3: Calculate the total moles in the solution
Total moles = moles of solute (urea) + moles of solvent (water)
\[ \text{Total moles} = n + N = 0.1 + 9.9 = 10 \, \text{mol} \]

### Step 4: Calculate the mole fraction of the solute (urea)
The mole fraction of the solute (urea) is given by:
\[ \text{Mole fraction of urea} = \frac{\text{moles of urea}}{\text{total moles}} = \frac{0.1}{10} = 0.01 \]

### Step 5: Use Raoult's Law to find the vapor pressure of the solution
According to Raoult's Law, the relative lowering of vapor pressure is equal to the mole fraction of the solute:
\[ \frac{P^0 - P_s}{P^0} = \text{Mole fraction of urea} \]
Substituting the mole fraction:
\[ \frac{P^0 - P_s}{P^0} = 0.01 \]

### Step 6: Rearranging to find the vapor pressure of the solution
Rearranging the equation gives:
\[ P^0 - P_s = 0.01 P^0 \]
\[ P_s = P^0 - 0.01 P^0 = 0.99 P^0 \]

### Conclusion
The vapor pressure of the solution (P_s) is:
\[ P_s = 0.99 P^0 \]