One mole of non-volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that water of

Options

A. `2//3`
B. `1//3`
C. `1//2`
D. `3//2`

Explanation

To solve the problem of finding the vapor pressure of a solution relative to that of pure water when one mole of a non-volatile solute is dissolved in two moles of water, we can use Raoult's Law. Here are the steps to arrive at the solution:

### Step 1: Identify the components
- We have 1 mole of a non-volatile solute.
- We have 2 moles of water (the solvent).

### Step 2: Calculate the total number of moles in the solution
Total moles = moles of solute + moles of solvent
Total moles = 1 mole (solute) + 2 moles (water) = 3 moles

### Step 3: Calculate the mole fraction of the solvent (water)
Mole fraction of water (solvent) = moles of water / total moles
Mole fraction of water = 2 moles / 3 moles = 2/3

### Step 4: Use Raoult's Law to find the vapor pressure of the solution
According to Raoult's Law:
- The vapor pressure of the solution (P) is given by:
\[ P = P_0 \times \text{(mole fraction of solvent)} \]
Where \( P_0 \) is the vapor pressure of pure solvent (water).

### Step 5: Substitute the values
Assuming \( P_0 \) is the vapor pressure of pure water, we can write:
\[ P = P_0 \times \frac{2}{3} \]

### Step 6: Calculate the relative vapor pressure
To find the relative vapor pressure of the solution to that of pure water, we take the ratio:
\[ \frac{P}{P_0} = \frac{P_0 \times \frac{2}{3}}{P_0} = \frac{2}{3} \]

### Final Answer
The vapor pressure of the solution relative to that of pure water is \( \frac{2}{3} \).

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