When a graph is plotted between log `x//m` and log p, it is straight line with an angle `45^(@)` and intercept `0.3010` on `y-` axis . If initial pressure is `0.3` atm, what wil be the amount of gas adsorbed per gm of adsorbent `:`

Options

A. `0.4g`
B. `0.6g`
C. `0.8g`
D. `0.1g`

Explanation

To solve the problem, we need to analyze the information given and apply the relevant equations.

### Step-by-Step Solution:

1. **Understanding the Graph**:
The graph plotted is between \( \log \frac{x}{m} \) (where \( x \) is the amount of gas adsorbed and \( m \) is the mass of the adsorbent) and \( \log p \) (where \( p \) is the pressure). The graph is a straight line with an angle of \( 45^\circ \).

**Hint**: A \( 45^\circ \) angle indicates that the slope of the line is 1.

2. **Slope and Intercept**:
The slope of the line can be expressed as \( \frac{1}{n} \) where \( n \) is a constant related to the adsorption process. Since the angle is \( 45^\circ \), we have:
\[ \frac{1}{n} = \tan(45^\circ) = 1 \implies n = 1 \]

**Hint**: The slope of a line at \( 45^\circ \) is equal to 1.

3. **Intercept**:
The y-intercept of the graph is given as \( 0.3010 \). This can be expressed in logarithmic form:
\[ \log k = 0.3010 \]
To find \( k \), we take the antilog:
\[ k = 10^{0.3010} \approx 2 \]

**Hint**: Use the antilogarithm to convert from log to the actual value.

4. **Using the Adsorption Isotherm**:
The relationship can be expressed as:
\[ \frac{x}{m} = k p^n \]
Since \( n = 1 \), we can simplify this to:
\[ \frac{x}{m} = k p \]

5. **Substituting Values**:
We know \( k = 2 \) and the initial pressure \( p = 0.3 \) atm. Substituting these values into the equation gives:
\[ \frac{x}{m} = 2 \times 0.3 = 0.6 \]

**Hint**: Substitute the known values into the equation to find \( \frac{x}{m} \).

6. **Finding the Amount of Gas Adsorbed**:
The value \( \frac{x}{m} = 0.6 \) means that the amount of gas adsorbed \( x \) per gram of adsorbent \( m \) is:
\[ x = 0.6 \text{ grams} \]

**Hint**: The result \( x \) represents the total amount of gas adsorbed per gram of adsorbent.

### Final Answer:
The amount of gas adsorbed per gram of adsorbent is **0.6 grams**.