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 follow these steps:

### Step 1: Calculate the molar mass of N₂O
The molar mass of N₂O can be calculated as follows:
- Nitrogen (N) has an atomic mass of approximately 14 g/mol. Since there are 2 nitrogen atoms, their contribution is \(2 \times 14 = 28 \, \text{g/mol}\).
- Oxygen (O) has an atomic mass of approximately 16 g/mol.
- Therefore, the molar mass of N₂O is:
\[ \text{Molar mass of N₂O} = 28 + 16 = 44 \, \text{g/mol} \]

### Step 2: Convert 200 mg of N₂O to grams
We need to convert 200 mg to grams:
\[ 200 \, \text{mg} = 200 \times 10^{-3} \, \text{g} = 0.2 \, \text{g} \]

### Step 3: Calculate the number of moles of N₂O in 200 mg
Using the molar mass, we can calculate the number of moles of N₂O in 0.2 g:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{0.2}{44} \approx 0.004545 \, \text{moles} \]

### Step 4: Convert moles of N₂O to molecules
Using Avogadro's number (\(6.022 \times 10^{23}\) molecules/mol), we can find the total number of molecules in 0.2 g of N₂O:
\[ \text{Number of molecules} = 0.004545 \, \text{moles} \times 6.022 \times 10^{23} \approx 2.73 \times 10^{21} \, \text{molecules} \]

### Step 5: Calculate the number of molecules left after removing x molecules
We know that after removing \(x\) molecules, \(2.89 \times 10^{-3}\) moles of N₂O are left. We need to convert this to molecules:
\[ \text{Molecules left} = 2.89 \times 10^{-3} \, \text{moles} \times 6.022 \times 10^{23} \approx 1.74 \times 10^{21} \, \text{molecules} \]

### Step 6: Set up the equation to find x
The total number of molecules before removing \(x\) is \(2.73 \times 10^{21}\). After removing \(x\) molecules, we have:
\[ 2.73 \times 10^{21} - x = 1.74 \times 10^{21} \]

### Step 7: Solve for x
Rearranging the equation gives:
\[ x = 2.73 \times 10^{21} - 1.74 \times 10^{21} = 0.99 \times 10^{21} \]
This is approximately equal to \(10^{21}\) molecules.

### Final Answer
Thus, the value of \(x\) is approximately \(10^{21}\) molecules.

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