`3.06 L` of `H_(2)O` vapour is taken at a pressure of `1 atm` and `373 K` . It is now condensed to `H_(2)O(l)` at `300 K`. Calculate the approximate volume occupied by `H_(2)O(l) :`
( Assume density of liquid water at `300 K =1000 kg//m^(3))`

Options

A. `3.06L`
B. `1.8mL`
C. `1.8L`
D. `3.06mL`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We have:
- Volume of water vapor (H₂O) = 3.06 L
- Pressure (P) = 1 atm
- Temperature (T) = 373 K
- Density of liquid water at 300 K = 1000 kg/m³

### Step 2: Convert the volume of water vapor to m³
Since the density is given in kg/m³, we should convert the volume from liters to cubic meters:
\[ 1 \text{ L} = 0.001 \text{ m}^3 \]
Thus,
\[ 3.06 \text{ L} = 3.06 \times 0.001 \text{ m}^3 = 0.00306 \text{ m}^3 \]

### Step 3: Use the Ideal Gas Law to find the mass of water vapor
The Ideal Gas Law is given by:
\[ PV = nRT \]
Where:
- \( n \) = number of moles
- \( R \) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature in Kelvin

Rearranging to find the number of moles \( n \):
\[ n = \frac{PV}{RT} \]

Substituting the values:
- P = 1 atm
- V = 3.06 L
- R = 0.0821 L·atm/(K·mol)
- T = 373 K

Calculating \( n \):
\[ n = \frac{(1 \text{ atm})(3.06 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)})(373 \text{ K})} \]
\[ n = \frac{3.06}{30.703} \approx 0.0997 \text{ mol} \]

### Step 4: Calculate the mass of water vapor
The molar mass of water (H₂O) is approximately 18 g/mol. Thus, the mass \( m \) can be calculated as:
\[ m = n \times \text{molar mass} \]
\[ m = 0.0997 \text{ mol} \times 18 \text{ g/mol} \approx 1.7946 \text{ g} \]

### Step 5: Convert mass to kilograms
\[ m = 1.7946 \text{ g} = 0.0017946 \text{ kg} \]

### Step 6: Calculate the volume of liquid water using density
Using the formula for density:
\[ \text{Density} = \frac{\text{mass}}{\text{volume}} \]
Rearranging gives:
\[ \text{Volume} = \frac{\text{mass}}{\text{density}} \]

Substituting the values:
- Mass = 0.0017946 kg
- Density = 1000 kg/m³

Calculating volume:
\[ \text{Volume} = \frac{0.0017946 \text{ kg}}{1000 \text{ kg/m}^3} = 0.0000017946 \text{ m}^3 \]
Converting to milliliters (1 m³ = 1,000,000 mL):
\[ \text{Volume} = 0.0000017946 \text{ m}^3 \times 1,000,000 \text{ mL/m}^3 \approx 1.7946 \text{ mL} \]

### Step 7: Round the answer
The approximate volume occupied by the liquid water is:
\[ \text{Volume} \approx 1.8 \text{ mL} \]

### Final Answer
The approximate volume occupied by \( H_2O(l) \) at 300 K is **1.8 mL**.

---