If `10^(-4) dm^(3)` of water is introduced into a `1.0 dm^(3)` flask to `300K` how many moles of water are in the vapour phase when equilibrium is established ? (Given vapour pressure of `H_(2)O` at `300K` is `3170Pa R =8.314 JK^(-1) mol^(-1))` .

Options

A. `1.27xx10^(-3)"mole"`
B. `5.56xx10^(-3)"mole"`
C. `1.53xx10^(-2)"mole"`
D. `4.46xx10^(-2)"mole"`

Explanation

To solve the problem of determining how many moles of water are in the vapor phase when equilibrium is established in a 1.0 dm³ flask at 300 K, we can follow these steps:

### Step 1: Understand the Given Information
- Volume of the flask (V) = 1.0 dm³ = 1.0 x 10⁻³ m³ (conversion to cubic meters)
- Volume of water introduced = 10⁻⁴ dm³ = 10⁻⁷ m³
- Temperature (T) = 300 K
- Vapor pressure of water at 300 K (P) = 3170 Pa
- Ideal gas constant (R) = 8.314 J/(K·mol)

### Step 2: Use the Ideal Gas Law
The number of moles of vapor (n) can be calculated using the ideal gas equation:
\[ PV = nRT \]

Rearranging the equation to solve for n gives:
\[ n = \frac{PV}{RT} \]

### Step 3: Substitute the Values
Now, substitute the values into the equation:
- P = 3170 Pa
- V = 1.0 x 10⁻³ m³
- R = 8.314 J/(K·mol)
- T = 300 K

So,
\[ n = \frac{(3170 \, \text{Pa}) \times (1.0 \times 10^{-3} \, \text{m}^3)}{(8.314 \, \text{J/(K·mol)}) \times (300 \, \text{K})} \]

### Step 4: Calculate the Numerator and Denominator
Calculating the numerator:
\[ 3170 \, \text{Pa} \times 1.0 \times 10^{-3} \, \text{m}^3 = 3.170 \, \text{J} \]

Calculating the denominator:
\[ 8.314 \, \text{J/(K·mol)} \times 300 \, \text{K} = 2494.2 \, \text{J/mol} \]

### Step 5: Calculate the Number of Moles
Now, substitute the calculated values back into the equation for n:
\[ n = \frac{3.170 \, \text{J}}{2494.2 \, \text{J/mol}} \]

Calculating n gives:
\[ n \approx 0.00127 \, \text{mol} \]
or
\[ n \approx 1.27 \times 10^{-3} \, \text{mol} \]

### Conclusion
The number of moles of water in the vapor phase when equilibrium is established is approximately **1.27 x 10⁻³ moles**.

---