Which of the following is not correct regarding 14 gram of carbon monoxide?

Options

A. It corresponds to `0*5` mole of CO
B. It occupies `2*24` litres at S.T.P.
C. It corresponds to `3*01xx10^(23)` molecules of CO
D. It corresponds to same number of moles of `CO_(2)` and nitrogen (I) oxide gases

Explanation

To solve the question regarding carbon monoxide (CO) and determine which statement is not correct, we will analyze the information step by step.

### Step 1: Determine the Molar Mass of Carbon Monoxide (CO)
- The atomic mass of carbon (C) is approximately 12 g/mol.
- The atomic mass of oxygen (O) is approximately 16 g/mol.
- Therefore, the molar mass of CO = 12 g/mol (C) + 16 g/mol (O) = 28 g/mol.

### Step 2: Calculate the Number of Moles in 14 grams of CO
- To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
- Substituting the values:
\[ \text{Number of moles of CO} = \frac{14 \text{ g}}{28 \text{ g/mol}} = 0.5 \text{ moles} \]

### Step 3: Evaluate the Given Options
Now, we will evaluate the statements provided in the options to identify which one is incorrect.

1. **Option A**: "Half mole of CO" - This is correct since we calculated that 14 grams of CO corresponds to 0.5 moles.

2. **Option B**: "CO is 22.4 L" - This statement is incorrect. The volume of 1 mole of any gas at standard temperature and pressure (STP) is 22.4 L. Since we have 0.5 moles of CO, the volume would be:
\[ \text{Volume} = 0.5 \text{ moles} \times 22.4 \text{ L/mole} = 11.2 \text{ L} \]
Therefore, stating that CO is 22.4 L is incorrect.

3. **Option C**: "CO is 6.0 x 10^23 molecules" - This is correct. The number of molecules in 1 mole is Avogadro's number (approximately \(6.022 \times 10^{23}\)). For 0.5 moles:
\[ \text{Number of molecules} = 0.5 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} \approx 3.01 \times 10^{23} \text{ molecules} \]
This statement is not directly stated but is inferred to be correct.

4. **Option D**: "The number of molecules of CO2 and NO2 is the same" - This statement is correct as it refers to stoichiometric relationships in reactions involving CO.

### Conclusion
The statement that is not correct regarding 14 grams of carbon monoxide is **Option B**: "CO is 22.4 L".