`CO_(2)` gas adsorbs on charcoal following Freundlich adsorption isotherm. For a given amount of charcoal, the mass of `CO_(2)` adsorbed becomes 64 times when the pressure of `CO_(2)` is doubled.
The value of n in the Freundlich isotherm equation is ________ ` xx 10^(-2)`. (Round off to the Nearest Integer)

Answer

To solve the problem, we will use the Freundlich adsorption isotherm equation, which is given by:

\[ \frac{x}{m} = k p^{\frac{1}{n}} \]

where:
- \( x \) is the mass of the gas adsorbed,
- \( m \) is the mass of the adsorbent (charcoal),
- \( k \) is a constant,
- \( p \) is the pressure of the gas,
- \( n \) is a constant that indicates the adsorption intensity.

### Step 1: Set up the initial conditions
Let:
- \( x_1 \) be the mass of \( CO_2 \) adsorbed at pressure \( p_1 \).
- \( x_2 \) be the mass of \( CO_2 \) adsorbed at pressure \( p_2 \).

According to the problem, when the pressure is doubled, the mass of \( CO_2 \) adsorbed becomes 64 times the initial amount:

\[ x_2 = 64 x_1 \]

### Step 2: Express the pressures
Since the pressure is doubled, we have:

\[ p_2 = 2 p_1 \]

### Step 3: Write the Freundlich isotherm equations for both states
Using the Freundlich isotherm for both conditions, we have:

1. For \( p_1 \):
\[ \frac{x_1}{m} = k p_1^{\frac{1}{n}} \]

2. For \( p_2 \):
\[ \frac{x_2}{m} = k (2 p_1)^{\frac{1}{n}} \]

### Step 4: Substitute \( x_2 \) in terms of \( x_1 \)
Substituting \( x_2 = 64 x_1 \) into the second equation gives:

\[ \frac{64 x_1}{m} = k (2 p_1)^{\frac{1}{n}} \]

### Step 5: Divide the two equations
Dividing the second equation by the first equation:

\[ \frac{64 x_1/m}{x_1/m} = \frac{k (2 p_1)^{\frac{1}{n}}}{k p_1^{\frac{1}{n}}} \]

This simplifies to:

\[ 64 = \frac{(2 p_1)^{\frac{1}{n}}}{p_1^{\frac{1}{n}}} \]

### Step 6: Simplify the right side
This can be simplified as follows:

\[ 64 = 2^{\frac{1}{n}} \]

### Step 7: Solve for \( n \)
Taking logarithms on both sides:

\[ \log(64) = \frac{1}{n} \log(2) \]

Since \( 64 = 2^6 \), we have:

\[ 6 \log(2) = \frac{1}{n} \log(2) \]

Dividing both sides by \( \log(2) \):

\[ 6 = \frac{1}{n} \]

Thus, we find:

\[ n = \frac{1}{6} \]

### Step 8: Convert \( n \) to the required format
To express \( n \) in the form of \( xx \times 10^{-2} \):

\[ n = \frac{1}{6} = 0.1666... \approx 0.17 \times 10^{-2} \]

### Step 9: Round off to the nearest integer
The nearest integer to \( 0.17 \times 10^{-2} \) is \( 17 \).

### Final Answer
The value of \( n \) in the Freundlich isotherm equation is:

\[ \boxed{17} \]