The number of `H_(2)O` molecules in a drop of water weighing 0.018 g is :

Options

A. `6.022 xx 10^(26)`
B. `6.022 xx 10^(23)`
C. `6.022 xx 10^(19)`
D. `6.022 xx 10^(20)`

Explanation

To find the number of \( H_2O \) molecules in a drop of water weighing 0.018 g, we can follow these steps:

### Step 1: Determine the molar mass of \( H_2O \)
The molecular formula for water is \( H_2O \), which consists of:
- 2 hydrogen (H) atoms, each with an atomic mass of approximately 1 g/mol.
- 1 oxygen (O) atom, with an atomic mass of approximately 16 g/mol.

Calculating the molar mass:
\[ \text{Molar mass of } H_2O = (2 \times 1) + (1 \times 16) = 2 + 16 = 18 \text{ g/mol} \]

### Step 2: Calculate the number of moles of water in 0.018 g
Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{0.018 \text{ g}}{18 \text{ g/mol}} = 0.001 \text{ moles} = 10^{-3} \text{ moles} \]

### Step 3: Use Avogadro's number to find the number of molecules
Avogadro's number (\( N_A \)) is approximately \( 6.022 \times 10^{23} \) molecules/mol. To find the number of molecules:
\[ \text{Number of molecules} = \text{Number of moles} \times N_A \]
Substituting the values:
\[ \text{Number of molecules} = 10^{-3} \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mol} \]
Calculating:
\[ \text{Number of molecules} = 6.022 \times 10^{20} \text{ molecules} \]

### Final Answer
The number of \( H_2O \) molecules in a drop of water weighing 0.018 g is approximately \( 6.022 \times 10^{20} \) molecules.

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