1 ml of gaseous aliphatic compound `C_(n) H_(3n) O_(m)` is completely burnt in an excess of `O_(2)` . The contraction in volume is

Options

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

Answer

To solve the problem of finding the contraction in volume when 1 ml of the gaseous aliphatic compound \( C_nH_{3n}O_m \) is completely burnt in excess \( O_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the General Reaction**:
The general combustion reaction for an aliphatic compound \( C_xH_yO_z \) can be represented as:
\[ C_xH_yO_z + O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O \]
Here, \( x \) is the number of carbon atoms, \( y \) is the number of hydrogen atoms, and \( z \) is the number of oxygen atoms.

2. **Substitute the Given Compound**:
For the compound \( C_nH_{3n}O_m \):
- \( x = n \)
- \( y = 3n \)
- \( z = m \)

3. **Calculate the Required Moles of \( O_2 \)**:
The moles of \( O_2 \) required for the reaction can be calculated using the formula:
\[ O_2 = x + \frac{y}{4} - \frac{z}{2} \]
Substituting the values:
\[ O_2 = n + \frac{3n}{4} - \frac{m}{2} \]
Simplifying this gives:
\[ O_2 = n + 0.75n - 0.5m = 1.75n - 0.5m \]

4. **Calculate the Total Moles of Reactants**:
The total moles of gaseous reactants (1 ml of the compound + moles of \( O_2 \)) is:
\[ \text{Total moles of reactants} = 1 + (1.75n - 0.5m) \]
This simplifies to:
\[ \text{Total moles of reactants} = 1 + 1.75n - 0.5m \]

5. **Calculate the Total Moles of Products**:
The products of the combustion will be \( CO_2 \) and \( H_2O \). The moles of products will be:
\[ \text{Total moles of products} = n + \frac{3n}{4} = n + 0.75n = 1.75n \]

6. **Determine the Contraction in Volume**:
The contraction in volume is given by the difference in moles of reactants and products:
\[ \text{Contraction in volume} = \text{Total moles of reactants} - \text{Total moles of products} \]
Substituting the values:
\[ \text{Contraction in volume} = (1 + 1.75n - 0.5m) - 1.75n \]
This simplifies to:
\[ \text{Contraction in volume} = 1 - 0.5m \]

### Final Answer:
The contraction in volume when 1 ml of the gaseous aliphatic compound \( C_nH_{3n}O_m \) is completely burnt in excess \( O_2 \) is:
\[ \text{Contraction in volume} = 1 - 0.5m \]