The total number of electrons in 2.0 g of `D_(2)O` to that in 1.8 g of `H_(2)O`

Options

A. Double
B. Same
C. Triple
D. One fourth

Answer

To solve the problem of finding the total number of electrons in 2.0 g of \( D_2O \) compared to 1.8 g of \( H_2O \), we will follow these steps:

### Step 1: Determine the molecular weight of \( D_2O \)
- Deuterium (\( D \)) has an atomic mass of approximately 2 g/mol.
- The molecular weight of \( D_2O \) can be calculated as follows:
\[ \text{Molecular weight of } D_2O = (2 \times 2) + 16 = 4 + 16 = 20 \text{ g/mol} \]

### Step 2: Calculate the number of moles of \( D_2O \)
- Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{Weight}}{\text{Molecular weight}} \]
- For \( D_2O \):
\[ \text{Number of moles of } D_2O = \frac{2.0 \text{ g}}{20 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 3: Determine the molecular weight of \( H_2O \)
- Hydrogen (\( H \)) has an atomic mass of approximately 1 g/mol.
- The molecular weight of \( H_2O \) can be calculated as follows:
\[ \text{Molecular weight of } H_2O = (2 \times 1) + 16 = 2 + 16 = 18 \text{ g/mol} \]

### Step 4: Calculate the number of moles of \( H_2O \)
- Using the formula for moles:
\[ \text{Number of moles of } H_2O = \frac{1.8 \text{ g}}{18 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 5: Calculate the total number of electrons in \( D_2O \)
- Each \( D_2O \) molecule contains 10 electrons (2 from deuterium and 8 from oxygen).
- Therefore, for 0.1 moles of \( D_2O \):
\[ \text{Total electrons in } D_2O = 0.1 \text{ moles} \times 10 \text{ electrons/molecule} \times N_A \]
where \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \)).
\[ = 0.1 \times 10 \times N_A = 1.0 \times N_A \]

### Step 6: Calculate the total number of electrons in \( H_2O \)
- Each \( H_2O \) molecule also contains 10 electrons (2 from hydrogen and 8 from oxygen).
- Therefore, for 0.1 moles of \( H_2O \):
\[ \text{Total electrons in } H_2O = 0.1 \text{ moles} \times 10 \text{ electrons/molecule} \times N_A \]
\[ = 0.1 \times 10 \times N_A = 1.0 \times N_A \]

### Step 7: Compare the total number of electrons
- The total number of electrons in both \( D_2O \) and \( H_2O \) is the same:
\[ \text{Total electrons in } D_2O = \text{Total electrons in } H_2O = 1.0 \times N_A \]

### Conclusion
The total number of electrons in 2.0 g of \( D_2O \) is equal to that in 1.8 g of \( H_2O \).

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