The weight of `2.01xx10^(23)` molecules of `CO` is-

Options

A. `9.3 g`
B. `7.2g`
C. `1.2g`
D. `3g`

Answer

To find the weight of \(2.01 \times 10^{23}\) molecules of carbon monoxide (CO), we can follow these steps:

### Step 1: Determine the number of moles
We know that one mole of any substance contains \(6.02 \times 10^{23}\) molecules (Avogadro's number). To find the number of moles (\(n\)) in \(2.01 \times 10^{23}\) molecules of CO, we can use the formula:

\[ n = \frac{\text{Number of molecules}}{\text{Avogadro's number}} = \frac{2.01 \times 10^{23}}{6.02 \times 10^{23}} \]

### Step 2: Calculate the number of moles
Now, we can perform the calculation:

\[ n = \frac{2.01}{6.02} \approx 0.3337 \text{ moles} \]

### Step 3: Find the molar mass of CO
The molar mass of carbon monoxide (CO) can be calculated by adding the atomic masses of carbon (C) and oxygen (O):

- Atomic mass of C = 12 g/mol
- Atomic mass of O = 16 g/mol

Thus, the molar mass of CO is:

\[ \text{Molar mass of CO} = 12 + 16 = 28 \text{ g/mol} \]

### Step 4: Calculate the mass of CO
Now that we have the number of moles and the molar mass, we can find the mass (\(m\)) of \(0.3337\) moles of CO using the formula:

\[ m = n \times \text{Molar mass} \]

Substituting the values:

\[ m = 0.3337 \text{ moles} \times 28 \text{ g/mol} \approx 9.34 \text{ grams} \]

### Final Answer
The weight of \(2.01 \times 10^{23}\) molecules of CO is approximately **9.34 grams**.

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