When x molecules are removed from 200 mg of `N_(2)O`. `2.89 xx 10^(-3)` moles of `N_(2)O` are left. x will be

Options

A. `10^(20)` molecules
B. `10^(10)` molecules
C. 21 molecules
D. `10^(21)` molecules

Answer

To solve the problem, we need to find the value of \( x \), which represents the number of molecules removed from 200 mg of \( N_2O \). We know that after removing \( x \) molecules, \( 2.89 \times 10^{-3} \) moles of \( N_2O \) are left.

### Step-by-Step Solution:

1. **Convert the mass of \( N_2O \) to grams**:
\[ 200 \text{ mg} = 0.200 \text{ g} \]

2. **Calculate the molar mass of \( N_2O \)**:
- The molar mass of nitrogen (\( N \)) is approximately 14 g/mol.
- The molar mass of oxygen (\( O \)) is approximately 16 g/mol.
\[ \text{Molar mass of } N_2O = (2 \times 14) + 16 = 28 + 16 = 44 \text{ g/mol} \]

3. **Calculate the number of moles of \( N_2O \) in 0.200 g**:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{0.200 \text{ g}}{44 \text{ g/mol}} \approx 0.004545 \text{ moles} \]

4. **Convert the number of moles to molecules**:
- Using Avogadro's number (\( 6.022 \times 10^{23} \) molecules/mol):
\[ \text{Number of molecules} = 0.004545 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mol} \approx 2.740 \times 10^{21} \text{ molecules} \]

5. **Determine the number of molecules left after removing \( x \) molecules**:
- We know that after removing \( x \) molecules, \( 2.89 \times 10^{-3} \) moles are left.
\[ \text{Moles left} = 2.89 \times 10^{-3} \text{ moles} \]
\[ \text{Number of molecules left} = 2.89 \times 10^{-3} \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mol} \approx 1.736 \times 10^{21} \text{ molecules} \]

6. **Calculate the number of molecules removed \( x \)**:
\[ x = \text{Initial number of molecules} - \text{Number of molecules left} \]
\[ x = 2.740 \times 10^{21} - 1.736 \times 10^{21} \approx 1.004 \times 10^{21} \text{ molecules} \]

### Final Answer:
The value of \( x \) is approximately \( 1.004 \times 10^{21} \) molecules.