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 of `N_2` `(iii)72.41% "by mass of `N_(2)`

Options

A. only i) is 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, we will follow these steps:

### Step 1: Assume the weight of air
Let's assume the weight of air is 100 g. This simplifies our calculations.

### Step 2: Define the weights of nitrogen and oxygen
Let the weight of nitrogen (N₂) in air be \( X \) grams. Then, the weight of oxygen (O₂) in air will be \( 100 - X \) grams.

### Step 3: Calculate the number of moles of nitrogen and oxygen
The number of moles can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{Weight}}{\text{Molecular weight}} \]
- For nitrogen (N₂), the molecular weight is 28 g/mol:
\[ \text{Moles of } N_2 = \frac{X}{28} \]
- For oxygen (O₂), the molecular weight is 32 g/mol:
\[ \text{Moles of } O_2 = \frac{100 - X}{32} \]

### Step 4: Set up the equation for average molecular mass
The average molecular mass of air is given as 29 g/mol. We can set up the equation:
\[ 29 = \frac{100}{\left(\frac{X}{28} + \frac{100 - X}{32}\right)} \]

### Step 5: Simplify the equation
Cross-multiply to eliminate the fraction:
\[ 29 \left(\frac{X}{28} + \frac{100 - X}{32}\right) = 100 \]
This expands to:
\[ 29 \left(\frac{X}{28}\right) + 29 \left(\frac{100 - X}{32}\right) = 100 \]

### Step 6: Solve for X
Multiply through by the least common multiple (LCM) of 28 and 32 (which is 896):
\[ 29 \cdot 32X + 29 \cdot 28(100 - X) = 100 \cdot 896 \]
This simplifies to:
\[ 928X + 81200 - 812X = 89600 \]
Combine like terms:
\[ 116X = 8400 \]
Thus,
\[ X = \frac{8400}{116} \approx 72.41 \text{ grams} \]

### Step 7: Calculate the weight of oxygen
Now, we can find the weight of oxygen:
\[ \text{Weight of } O_2 = 100 - X = 100 - 72.41 = 27.59 \text{ grams} \]

### Step 8: Calculate the mass percentage of nitrogen
The mass percentage of nitrogen is given by:
\[ \text{Mass \% of } N_2 = \left(\frac{72.41}{100}\right) \times 100 = 72.41\% \]

### Step 9: Calculate the number of moles of nitrogen and oxygen
- 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 10: Calculate total moles and mole percentage of nitrogen
Total moles:
\[ \text{Total moles} = 2.59 + 0.86 = 3.45 \]
Mole percentage of nitrogen:
\[ \text{Mole \% of } N_2 = \left(\frac{2.59}{3.45}\right) \times 100 \approx 75\% \]

### Conclusion
The correct options regarding the composition of air are:
- (ii) 75% by moles of N₂
- (iii) 72.41% by mass of N₂