3 gms of urea is dissolved in 45 gms of H2O. The relative lowering in vapour pressure is :

Explanation

To solve the problem of finding the relative lowering in vapor pressure when 3 grams of urea is dissolved in 45 grams of water, we can follow these steps:

### Step 1: Understand the Concept
Relative lowering of vapor pressure is given by the formula:
\[ \text{Relative lowering} = \frac{P_0 - P}{P_0} = \text{mole fraction of solute} \]
where \(P_0\) is the vapor pressure of the pure solvent and \(P\) is the vapor pressure of the solution.

### Step 2: Calculate Moles of Urea
First, we need to calculate the number of moles of urea. The formula to calculate moles is:
\[ \text{Moles} = \frac{\text{Given mass}}{\text{Molecular mass}} \]
For urea:
- Given mass = 3 grams
- Molecular mass of urea = 60 g/mol

Calculating the moles of urea:
\[ \text{Moles of urea} = \frac{3 \text{ g}}{60 \text{ g/mol}} = 0.05 \text{ moles} \]

### Step 3: Calculate Moles of Water
Next, we calculate the number of moles of water using the same formula:
- Given mass of water = 45 grams
- Molecular mass of water = 18 g/mol

Calculating the moles of water:
\[ \text{Moles of water} = \frac{45 \text{ g}}{18 \text{ g/mol}} = 2.5 \text{ moles} \]

### Step 4: Calculate Mole Fraction of Urea
The mole fraction of the solute (urea) is given by:
\[ \text{Mole fraction of urea} = \frac{\text{Moles of urea}}{\text{Moles of urea} + \text{Moles of water}} \]
Substituting the values:
\[ \text{Mole fraction of urea} = \frac{0.05}{0.05 + 2.5} = \frac{0.05}{2.55} \approx 0.0196 \]

### Step 5: Calculate Relative Lowering of Vapor Pressure
Since the relative lowering of vapor pressure is equal to the mole fraction of the solute:
\[ \text{Relative lowering of vapor pressure} \approx 0.0196 \]

### Final Answer
The relative lowering in vapor pressure is approximately \(0.0196\).

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