A sample of the liquid `H_(2)O` at 18.0 g is injected into an evacuated 7.6 L flask maintained at `27.0^(@)C`. If vapour pressure of `H_(2)O` at `27.0^(@)C` is 24.63 mm Hg, what weight percentage of the water will be vaporised when the system comes to equilibrium? Assume water vapors behaves as an ideal gas. The volume occupied by the liquid water is negligible compared to the volume of the container:

Options

A. 1%
B. 10%
C. 18%
D. 20%

Answer

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Understand the Given Data
We have:
- Mass of water (H₂O) = 18.0 g
- Volume of the flask = 7.6 L
- Temperature = 27.0 °C
- Vapor pressure of water at 27.0 °C = 24.63 mm Hg

### Step 2: Convert Vapor Pressure to Atmospheres
To use the ideal gas law, we need to convert the vapor pressure from mm Hg to atm.
\[ \text{Pressure in atm} = \frac{24.63 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} \approx 0.0324 \, \text{atm} \]

### Step 3: Convert Temperature to Kelvin
Convert the temperature from Celsius to Kelvin:
\[ T(K) = 27 + 273 = 300 \, K \]

### Step 4: Use the Ideal Gas Law to Find Moles of Water Vapor
Using the ideal gas equation \(PV = nRT\), we can rearrange it to find the number of moles (n):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(0.0324 \, \text{atm}) \times (7.6 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (300 \, K)} \]

Calculating \(n\):
\[ n = \frac{0.24624}{24.63} \approx 0.01 \, \text{mol} \]

### Step 5: Calculate the Mass of Vaporized Water
Now, we can find the mass of the vaporized water using the molar mass of water (18 g/mol):
\[ \text{Mass of vaporized water} = n \times \text{Molar mass} = 0.01 \, \text{mol} \times 18 \, \text{g/mol} = 0.18 \, \text{g} \]

### Step 6: Calculate the Weight Percentage of Vaporized Water
To find the weight percentage of the vaporized water:
\[ \text{Weight percentage} = \left( \frac{\text{mass of vaporized water}}{\text{initial mass of water}} \right) \times 100 \]
Substituting the values:
\[ \text{Weight percentage} = \left( \frac{0.18 \, \text{g}}{18.0 \, \text{g}} \right) \times 100 = 1\% \]

### Final Answer
The weight percentage of the water that will be vaporized when the system comes to equilibrium is **1%**.

---