If average molecular mass of air is 29, then assuming `N_(2)` and `O_(2)` gases are there, which option are correct regarding composition of are?
(i)`75% "by mass of" N_(2)" "(ii)75% "by moles "N_(2)" "(iii)72.41% "by mass of" N_(2)`

Options

A. only (i) is are correct
B. Only (ii) is correct
C. both (ii) and (iii) are correct
D. both (i) and (ii) are correct

Answer

To solve the problem regarding the composition of air based on the average molecular mass, we will follow these steps:

### Step 1: Assume the Weight of Air
Assume the weight of air is 100 g for simplicity. This will help us in calculating the mass percentages easily.

### Step 2: Define Variables for Nitrogen and Oxygen
Let the weight of nitrogen (N₂) in air be \( X \) grams. Consequently, the weight of oxygen (O₂) in air will be \( 100 - X \) grams.

### Step 3: Calculate the Number of Moles
Using the molecular weights:
- Molecular weight of nitrogen (N₂) = 28 g/mol
- Molecular weight of oxygen (O₂) = 32 g/mol

The number of moles of nitrogen can be calculated as:
\[ \text{Number of moles of } N_2 = \frac{X}{28} \]

The number of moles of oxygen can be calculated as:
\[ \text{Number of moles of } O_2 = \frac{100 - X}{32} \]

### Step 4: Set Up the Average Molecular Mass Equation
The average molecular mass of air is given as 29 g/mol. The equation for average molecular mass is:
\[ \text{Average Molecular Mass} = \frac{\text{Total Mass}}{\text{Total Moles}} \]
Substituting the values:
\[ 29 = \frac{100}{\left(\frac{X}{28} + \frac{100 - X}{32}\right)} \]

### Step 5: Solve for \( X \)
Cross-multiply and simplify the equation:
\[ 29 \left(\frac{X}{28} + \frac{100 - X}{32}\right) = 100 \]

This expands to:
\[ 29 \left(\frac{X}{28}\right) + 29 \left(\frac{100 - X}{32}\right) = 100 \]

Now, multiply through by the least common multiple (LCM) of 28 and 32 (which is 896):
\[ 29 \cdot 32X + 29 \cdot 28(100 - X) = 100 \cdot 896 \]

Solving this will yield:
\[ 32X + 28(100 - X) = 100 \cdot \frac{896}{29} \]
After calculations, we find \( X \approx 72.41 \) grams.

### Step 6: Calculate the Mass of Oxygen
Now, calculate the mass of oxygen:
\[ \text{Weight of } O_2 = 100 - X = 100 - 72.41 = 27.59 \text{ grams} \]

### Step 7: Calculate Mass Percentages
- Mass percentage of nitrogen:
\[ \text{Mass \% of } N_2 = \left(\frac{72.41}{100}\right) \times 100 = 72.41\% \]
- Mass percentage of oxygen:
\[ \text{Mass \% of } O_2 = \left(\frac{27.59}{100}\right) \times 100 = 27.59\% \]

### Step 8: Calculate Moles for Mole Percentage
Now, calculate the number of moles:
- Moles of nitrogen:
\[ \text{Moles of } N_2 = \frac{72.41}{28} \approx 2.59 \]
- Moles of oxygen:
\[ \text{Moles of } O_2 = \frac{27.59}{32} \approx 0.86 \]

### Step 9: Total Moles
Total moles:
\[ \text{Total moles} = 2.59 + 0.86 = 3.45 \]

### Step 10: Calculate Mole Percentages
- Mole percentage of nitrogen:
\[ \text{Mole \% of } N_2 = \left(\frac{2.59}{3.45}\right) \times 100 \approx 75\% \]

### Conclusion
Based on the calculations:
- (i) 75% by mass of N₂ is incorrect (it is 72.41%).
- (ii) 75% by moles of N₂ is correct.
- (iii) 72.41% by mass of N₂ is correct.

### Final Answer
Correct options are (ii) and (iii).

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