1 mol of a gaseous aliphatic compound `C_nH_(3n)O_m` is completely burnt in an excess of oxygen. The contraction in volume in (assume water gets condensed out ):

Options

A. `(1+(1)/(2)n-(3)/(4)m)`
B. `(1+(3)/(4)n-(1)/(4)m)`
C. `(1+(1)/(2)n-(3)/(4)m)`
D. `(1+(3)/(4)n-(1)/(2)m)`

Explanation

To solve the problem of finding the contraction in volume when 1 mole of the gaseous aliphatic compound \( C_nH_{3n}O_m \) is completely burnt in excess oxygen, we can follow these steps:

### Step 1: Determine the products of combustion
When a hydrocarbon burns in oxygen, it produces carbon dioxide (CO₂) and water (H₂O). The general combustion reaction can be represented as follows:
\[ C_nH_{3n}O_m + O_2 \rightarrow CO_2 + H_2O \]

### Step 2: Write the balanced equation for combustion
From the combustion of \( C_nH_{3n}O_m \), we can derive the moles of products formed:
- Each mole of carbon (\( C \)) produces 1 mole of \( CO_2 \).
- Each 2 moles of hydrogen (\( H \)) produce 1 mole of \( H_2O \).

Thus, the combustion of \( C_nH_{3n}O_m \) will produce:
- \( n \) moles of \( CO_2 \)
- \( \frac{3n}{2} \) moles of \( H_2O \)

### Step 3: Calculate the moles of oxygen required
To find the moles of oxygen (\( O_2 \)) required for the reaction, we can use the following formula:
\[ \text{Moles of } O_2 = \text{Moles of } CO_2 + \frac{1}{2} \text{Moles of } H_2O - \frac{1}{2} \text{Moles of } O \]
Substituting the values:
\[ \text{Moles of } O_2 = n + \frac{3n}{4} - \frac{m}{2} \]

### Step 4: Total moles before and after combustion
Initially, we have:
- 1 mole of \( C_nH_{3n}O_m \) (the reactant)
- Moles of \( O_2 \) consumed (which we calculated)

The total moles before combustion:
\[ \text{Total moles before} = 1 + \text{Moles of } O_2 \]

The total moles after combustion:
\[ \text{Total moles after} = n + \frac{3n}{2} \]

### Step 5: Calculate the contraction in volume
The contraction in volume can be calculated as:
\[ \text{Contraction} = \text{Total moles before} - \text{Total moles after} \]
Substituting the values:
\[ \text{Contraction} = \left(1 + n + \frac{3n}{4} - \frac{m}{2}\right) - \left(n + \frac{3n}{2}\right) \]
Simplifying this expression gives:
\[ \text{Contraction} = 1 + \frac{3n}{4} - \frac{m}{2} - n - \frac{3n}{2} \]
\[ = 1 + \frac{3n}{4} - n - \frac{3n}{2} - \frac{m}{2} \]
\[ = 1 + \frac{3n}{4} - \frac{4n}{4} - \frac{6n}{4} - \frac{m}{2} \]
\[ = 1 - \frac{7n}{4} - \frac{m}{2} \]

### Final Result
Thus, the contraction in volume when 1 mole of the compound \( C_nH_{3n}O_m \) is completely burnt in excess oxygen is:
\[ \text{Contraction} = 1 - \frac{7n}{4} - \frac{m}{2} \]