STATEMENT -1 : Equal weight of `.^(14)CO_(2)` and `NO_(2)` have same number of molecule.
STATEMENT -2 : `.^(14)CO_(2)` and `NO_(2)` have same molecular weight.
STATEMENT - 3 : Equal volume of `CO_(2)` and `NO_(2)` have same weight.

Options

A. TFF
B. TTF
C. FTT
D. FFT

Answer

To solve the question, we need to evaluate the three statements regarding the gases \(^{14}CO_2\) (carbon dioxide) and \(NO_2\) (nitrogen dioxide) one by one.

### Step 1: Evaluate Statement 1
**Statement 1**: Equal weight of \(^{14}CO_2\) and \(NO_2\) have the same number of molecules.

1. **Calculate Molecular Weights**:
- For \(^{14}CO_2\):
- Atomic mass of Carbon (C) = 14
- Atomic mass of Oxygen (O) = 16
- Molecular weight of \(^{14}CO_2 = 14 + 2 \times 16 = 14 + 32 = 46 \, \text{g/mol}\)
- For \(NO_2\):
- Atomic mass of Nitrogen (N) = 14
- Molecular weight of \(NO_2 = 14 + 2 \times 16 = 14 + 32 = 46 \, \text{g/mol}\)

2. **Calculate Number of Moles**:
- If we take equal weight \(X\) grams of both gases:
- Moles of \(^{14}CO_2 = \frac{X}{46}\)
- Moles of \(NO_2 = \frac{X}{46}\)

3. **Calculate Number of Molecules**:
- Number of molecules = Moles × Avogadro's number (\(N_A\)):
- For \(^{14}CO_2\): \(\frac{X}{46} \times N_A\)
- For \(NO_2\): \(\frac{X}{46} \times N_A\)

Since both gases have the same moles for equal weight, they will have the same number of molecules.

**Conclusion**: Statement 1 is **True**.

### Step 2: Evaluate Statement 2
**Statement 2**: \(^{14}CO_2\) and \(NO_2\) have the same molecular weight.

From our calculations in Step 1, we found that:
- Molecular weight of \(^{14}CO_2 = 46 \, \text{g/mol}\)
- Molecular weight of \(NO_2 = 46 \, \text{g/mol}\)

**Conclusion**: Statement 2 is **True**.

### Step 3: Evaluate Statement 3
**Statement 3**: Equal volume of \(CO_2\) and \(NO_2\) have the same weight.

1. **Calculate Moles in Equal Volume**:
- At STP, 1 mole of any gas occupies 22.4 L.
- For a volume \(V\) L, the number of moles for both gases:
- Moles of \(^{14}CO_2 = \frac{V}{22.4}\)
- Moles of \(NO_2 = \frac{V}{22.4}\)

2. **Calculate Weights**:
- Weight of \(^{14}CO_2 = \text{Moles} \times \text{Molecular Weight} = \frac{V}{22.4} \times 44\)
- Weight of \(NO_2 = \text{Moles} \times \text{Molecular Weight} = \frac{V}{22.4} \times 46\)

3. **Comparison**:
- Weight of \(^{14}CO_2 = \frac{V \times 44}{22.4}\)
- Weight of \(NO_2 = \frac{V \times 46}{22.4}\)

Since \(44 \neq 46\), the weights are different.

**Conclusion**: Statement 3 is **False**.

### Final Conclusion
- Statement 1: True
- Statement 2: True
- Statement 3: False

Thus, the correct option is **B: True, True, False**.