6.0 g of urea (molecular mass = 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. 1.10 `P^(@)`
C. 0.90 `P^(@)`
D. 0.99 `P^(@)`

Answer

To solve the problem, we need to determine the vapor pressure of a solution containing urea dissolved in water. Here are the steps to arrive at the solution:

### Step 1: Calculate the moles of urea
To find the number of moles of urea (solute), we can use the formula:

\[ \text{Moles of urea} = \frac{\text{mass of urea}}{\text{molecular mass of urea}} \]

Given:
- Mass of urea = 6.0 g
- Molecular mass of urea = 60 g/mol

Calculating the moles:

\[ \text{Moles of urea} = \frac{6.0 \, \text{g}}{60 \, \text{g/mol}} = 0.1 \, \text{moles} \]

### Step 2: Identify the moles of water (solvent)
The problem states that there are 9.9 moles of water.

### Step 3: Calculate the mole fraction of the solute (urea)
The mole fraction of the solute (urea) can be calculated using the formula:

\[ \text{Mole fraction of solute} (X_{\text{solute}}) = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]

Substituting the values:

\[ X_{\text{solute}} = \frac{0.1}{0.1 + 9.9} = \frac{0.1}{10.0} = 0.01 \]

### Step 4: Use Raoult's Law to find the vapor pressure of the solution
According to Raoult's Law, the relative lowering of vapor pressure is given by:

\[ \frac{P^0 - P_s}{P^0} = X_{\text{solute}} \]

Where:
- \(P^0\) = vapor pressure of pure water
- \(P_s\) = vapor pressure of the solution

Substituting the mole fraction of the solute:

\[ \frac{P^0 - P_s}{P^0} = 0.01 \]

### Step 5: Rearranging the equation to find \(P_s\)
Rearranging the equation gives:

\[ P^0 - P_s = 0.01 P^0 \]

Now, solving for \(P_s\):

\[ P_s = P^0 - 0.01 P^0 = 0.99 P^0 \]

### Final Answer
The vapor pressure of the solution is:

\[ P_s = 0.99 P^0 \]