What is the mass of a water molecule in gram ? How many molecules are present in one drop of pure water which weight 0.05 g ? If the same drop of water evaporates in one hour, calculate the number of molecules leaving the liquid surface per second.

Explanation

To solve the problem step by step, we will address each part of the question systematically.

### Step 1: Calculate the mass of a water molecule in grams.

1. **Identify the molar mass of water (H₂O)**:
- The molar mass of water is approximately 18 grams per mole.

2. **Use Avogadro's number**:
- Avogadro's number (NA) is \(6.022 \times 10^{23}\) molecules per mole.

3. **Calculate the mass of one water molecule**:
\[ \text{Mass of one water molecule} = \frac{\text{Molar mass of water}}{\text{Avogadro's number}} = \frac{18 \text{ g/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 2.99 \times 10^{-23} \text{ g} \]

### Step 2: Calculate the number of molecules in one drop of pure water weighing 0.05 g.

1. **Use the mass of the drop**:
- Given mass of the drop = 0.05 g.

2. **Calculate the number of moles in the drop**:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.05 \text{ g}}{18 \text{ g/mol}} \approx 0.00278 \text{ mol} \]

3. **Calculate the number of molecules**:
\[ \text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's number} = 0.00278 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \approx 1.67 \times 10^{21} \text{ molecules} \]

### Step 3: Calculate the number of molecules leaving the liquid surface per second if the drop evaporates in one hour.

1. **Convert the evaporation time to seconds**:
- 1 hour = 3600 seconds.

2. **Calculate the number of molecules evaporating per second**:
\[ \text{Molecules evaporating per second} = \frac{\text{Total number of molecules}}{\text{Time in seconds}} = \frac{1.67 \times 10^{21} \text{ molecules}}{3600 \text{ s}} \approx 4.64 \times 10^{17} \text{ molecules/s} \]

### Final Answers:
- The mass of a water molecule is approximately \(2.99 \times 10^{-23}\) grams.
- The number of molecules in one drop of pure water weighing 0.05 g is approximately \(1.67 \times 10^{21}\) molecules.
- The number of molecules leaving the liquid surface per second during evaporation is approximately \(4.64 \times 10^{17}\) molecules/s.