Plot of log against log P is a straight line inclined at an angle of `45^(@)`. When the pressure is 0.5 atm and Freundlich parameter ,K is 10, the amount of solute adsorbed per gram of adsorbent will be : (log 5=0.6990 )

Options

A. 5 g
B. 3 g
C. 6 g
D. 12 g

Explanation

To solve the problem, we will use the Freundlich adsorption isotherm equation and the information provided in the question. Here’s a step-by-step solution:

### Step 1: Write the Freundlich Adsorption Isotherm Equation
The Freundlich adsorption isotherm is given by the equation:
\[ \frac{X}{M} = K P^{\frac{1}{n}} \]
where:
- \(X\) is the amount of solute adsorbed,
- \(M\) is the mass of the adsorbent,
- \(K\) is the Freundlich constant,
- \(P\) is the pressure,
- \(n\) is the Freundlich exponent.

### Step 2: Take the Logarithm of Both Sides
Taking the logarithm of both sides gives:
\[ \log\left(\frac{X}{M}\right) = \frac{1}{n} \log P + \log K \]

### Step 3: Identify the Slope and Intercept
This equation can be compared to the equation of a straight line \(y = mx + c\), where:
- \(y = \log\left(\frac{X}{M}\right)\),
- \(x = \log P\),
- \(m = \frac{1}{n}\) (slope),
- \(c = \log K\) (intercept).

### Step 4: Determine the Value of \(n\)
Given that the plot of \(\log\left(\frac{X}{M}\right)\) against \(\log P\) is a straight line inclined at an angle of \(45^\circ\), we know that:
\[ \tan(45^\circ) = 1 \]
Thus, the slope \(m = \frac{1}{n} = 1\), which implies:
\[ n = 1 \]

### Step 5: Substitute the Given Values
Now we substitute the known values into the Freundlich equation. We know:
- \(K = 10\),
- \(P = 0.5 \, \text{atm}\).

Substituting these values into the logarithmic form:
\[ \log\left(\frac{X}{M}\right) = \frac{1}{1} \log(0.5) + \log(10) \]

### Step 6: Calculate \(\log(0.5)\)
Using the logarithmic property:
\[ \log(0.5) = \log\left(\frac{5}{10}\right) = \log 5 - \log 10 \]
Since \(\log 10 = 1\), we have:
\[ \log(0.5) = \log 5 - 1 \]

### Step 7: Substitute \(\log(0.5)\) Back into the Equation
Now substituting back:
\[ \log\left(\frac{X}{M}\right) = (\log 5 - 1) + 1 \]
This simplifies to:
\[ \log\left(\frac{X}{M}\right) = \log 5 \]

### Step 8: Take the Antilog
Taking the antilogarithm of both sides:
\[ \frac{X}{M} = 5 \]

### Step 9: Conclusion
Thus, the amount of solute adsorbed per gram of adsorbent is:
\[ X = 5 \, \text{grams} \]

### Final Answer
The amount of solute adsorbed per gram of adsorbent will be **5 grams**.

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