0.6 g urea is added to 360 g water. Calculate lowering in vapor pressure for this solution (given: vapour pressure of `H_(2)O` is 35 mm of Hg)

Options

A. `0.027 mm Hg`
B. `0.028 mm Hg`
C. `0.017 mm Hg`
D. `0.031 mm Hg`

Answer

To calculate the lowering in vapor pressure when 0.6 g of urea is added to 360 g of water, we can follow these steps:

### Step 1: Identify the given values
- Mass of urea (solute) = 0.6 g
- Mass of water (solvent) = 360 g
- Vapor pressure of pure water (P0) = 35 mm Hg

### Step 2: Calculate the number of moles of urea
To find the number of moles of urea, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of urea (NH2CONH2) is approximately 60 g/mol.

\[ \text{Moles of urea} = \frac{0.6 \text{ g}}{60 \text{ g/mol}} = 0.01 \text{ moles} \]

### Step 3: Calculate the number of moles of water
The molar mass of water (H2O) is approximately 18 g/mol.

\[ \text{Moles of water} = \frac{360 \text{ g}}{18 \text{ g/mol}} = 20 \text{ moles} \]

### Step 4: Calculate the total number of moles in the solution
\[ \text{Total moles} = \text{Moles of urea} + \text{Moles of water} = 0.01 + 20 = 20.01 \text{ moles} \]

### Step 5: Calculate the mole fraction of urea
The mole fraction of urea (X_urea) is given by:
\[ X_{\text{urea}} = \frac{\text{Moles of urea}}{\text{Total moles}} = \frac{0.01}{20.01} \]

Calculating this gives:
\[ X_{\text{urea}} \approx 0.00049975 \]

### Step 6: Calculate the lowering in vapor pressure
Using the formula for lowering in vapor pressure:
\[ \Delta P = P_0 \times X_{\text{urea}} \]
Substituting the values:
\[ \Delta P = 35 \text{ mm Hg} \times 0.00049975 \approx 0.017 \text{ mm Hg} \]

### Final Answer
The lowering in vapor pressure is approximately **0.017 mm Hg**.

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