In which case the number of water molecules is 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^-2` mol of water

Answer

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

### Step 1: Analyze the first case (0.00224 liters of water vapor at STP)
- At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters.
- Given volume = 0.00224 liters.
- Moles of water vapor = Volume / Molar Volume = 0.00224 L / 22.4 L/mol = 0.0001 moles.
- Number of water molecules = Moles × Avogadro's number = 0.0001 moles × 6.023 × 10²³ molecules/mole = 6.023 × 10²⁰ molecules.

### Step 2: Analyze the second case (0.18 g of water)
- Molar mass of water (H₂O) = 18 g/mol.
- Moles of water = Mass / Molar Mass = 0.18 g / 18 g/mol = 0.01 moles.
- Number of water molecules = Moles × Avogadro's number = 0.01 moles × 6.023 × 10²³ molecules/mole = 6.023 × 10²¹ molecules.

### Step 3: Analyze the third case (18 ml of water)
- Density of water = 1 g/ml, so 18 ml of water has a mass of 18 g.
- Moles of water = Mass / Molar Mass = 18 g / 18 g/mol = 1 mole.
- Number of water molecules = Moles × Avogadro's number = 1 mole × 6.023 × 10²³ molecules/mole = 6.023 × 10²³ molecules.

### Step 4: Analyze the fourth case (0.01 moles of water)
- Given directly as moles = 0.01 moles.
- Number of water molecules = Moles × Avogadro's number = 0.01 moles × 6.023 × 10²³ molecules/mole = 6.023 × 10²¹ molecules.

### Summary of Results
- Case 1: 6.023 × 10²⁰ molecules
- Case 2: 6.023 × 10²¹ molecules
- Case 3: 6.023 × 10²³ molecules
- Case 4: 6.023 × 10²¹ molecules

### Conclusion
The maximum number of water molecules is in **Case 3**, which contains **18 ml of water**.

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