When a graph is plotted between log x/m and log p, it is straight line with an angle `45^@` and intercept 0.6020 on y-axis.if initial pressure is 0.3 atm, what will be the amount of gas adsorbed per gram of adsorbent :

Options

A. 0.4
B. 1.2
C. 0.8
D. 0.1

Explanation

To solve the problem step by step, we will use the Freundlich Isotherm equation and the information provided in the question.

### Step 1: Understanding the Freundlich Isotherm
The Freundlich Isotherm is given by the equation:
\[ \frac{x}{m} = k \cdot p^{\frac{1}{n}} \]
where:
- \(x\) = mass of the gas adsorbed,
- \(m\) = mass of the adsorbent,
- \(p\) = pressure of the gas,
- \(k\) = Freundlich constant,
- \(n\) = constant related to the adsorption intensity.

### Step 2: Logarithmic Transformation
Taking logarithms of both sides, we get:
\[ \log\left(\frac{x}{m}\right) = \log(k) + \frac{1}{n} \log(p) \]
This can be rearranged to the form of a straight line \(y = mx + c\):
- \(y = \log\left(\frac{x}{m}\right)\)
- \(x = \log(p)\)
- Slope \(m = \frac{1}{n}\)
- Y-intercept \(c = \log(k)\)

### Step 3: Analyzing the Given Information
From the problem:
- The graph has a slope corresponding to an angle of \(45^\circ\), which means \(\tan(45^\circ) = 1\). Therefore, \(\frac{1}{n} = 1\) implies \(n = 1\).
- The y-intercept is given as \(0.6020\), which means:
\[ \log(k) = 0.6020 \implies k = 10^{0.6020} \approx 4 \]

### Step 4: Calculating the Amount of Gas Adsorbed
Using the Freundlich Isotherm equation with the values we have:
\[ \frac{x}{m} = k \cdot p^{\frac{1}{n}} \]
Substituting \(k = 4\), \(p = 0.3\), and \(n = 1\):
\[ \frac{x}{m} = 4 \cdot (0.3)^{1} \]
\[ \frac{x}{m} = 4 \cdot 0.3 = 1.2 \]

### Conclusion
The amount of gas adsorbed per gram of adsorbent is \(1.2\).

### Final Answer
The amount of gas adsorbed per gram of adsorbent is **1.2**.

---