In which case is the number of molecules of water maximum?

Options

A. 0.00224 L of water vapours at 1 atm and 273 K
B. 0.18 g of water
C. 18 mL of water
D. `10^(-3)` mol of water

Explanation

To determine in which case the number of molecules of water is maximum, we will analyze each option provided in the question step by step.

### Step 1: Analyze the first option
- **Given**: Volume of water vapor = 0.00224 liters, Pressure = 180 m, Temperature = 273 K.
- **Formula**: Number of moles (n) = Volume of gas (L) / 22.4 L (at STP).

Calculating the number of moles:
\[ n = \frac{0.00224 \, \text{L}}{22.4 \, \text{L}} = 0.0001 \, \text{moles} \]

### Step 2: Analyze the second option
- **Given**: Mass of water = 0.18 grams.
- **Molar mass of water (H2O)**: 2(1) + 16 = 18 g/mol.

Calculating the number of moles:
\[ n = \frac{0.18 \, \text{g}}{18 \, \text{g/mol}} = 0.01 \, \text{moles} \]

### Step 3: Analyze the third option
- **Given**: Volume of water = 18 mL.
- **Density of water**: 1 g/mL, so mass = 18 g.

Calculating the number of moles:
\[ n = \frac{18 \, \text{g}}{18 \, \text{g/mol}} = 1 \, \text{mole} \]

### Step 4: Analyze the fourth option
- **Given**: Number of moles = \( 10^{-4} \) moles.

### Step 5: Compare the number of moles from all options
1. First option: \( 0.0001 \) moles
2. Second option: \( 0.01 \) moles
3. Third option: \( 1 \) mole
4. Fourth option: \( 0.0001 \) moles

### Conclusion
The third option has the maximum number of moles, which is 1 mole. Since the number of molecules is directly proportional to the number of moles, the third option also has the maximum number of molecules of water.

**Final Answer**: The third option has the maximum number of molecules of water.

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