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In an adsorption experiment, a graph of ...

In an adsorption experiment, a graph of log(x/m) versus log P was found to be linear with a slope of 45°, and the the intercept is found to be 0.3010. The amount of gas adsorbed per gram charcoal under a pressure of 0.8 atm is

A

1.2

B

1.4

C

1.6

D

1.8

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The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Understand the Freundlich Adsorption Isotherm The Freundlich adsorption isotherm relates the amount of gas adsorbed per gram of adsorbent (x/m) to the pressure (P) of the gas. The equation can be expressed as: \[ \frac{x}{m} = k \cdot P^{\frac{1}{n}} \] where: - \( x \) = amount of gas adsorbed - \( m \) = mass of the adsorbent - \( k \) = Freundlich constant - \( P \) = pressure - \( n \) = constant related to the adsorption intensity ### Step 2: Take the Logarithm of Both Sides Taking the logarithm of both sides gives us: \[ \log\left(\frac{x}{m}\right) = \log(k) + \frac{1}{n} \log(P) \] ### Step 3: Identify the Slope and Intercept This equation resembles 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) From the problem, we know: - The slope is 45°, which means \( \tan(45°) = 1 \). Therefore, \( \frac{1}{n} = 1 \) and thus \( n = 1 \). - The intercept is given as 0.3010, so \( \log(k) = 0.3010 \). ### Step 4: Calculate the Value of k To find \( k \), we can convert the logarithmic value back to its exponential form: \[ k = 10^{0.3010} \] We know that \( 10^{0.3010} \) is equal to 2 (since \( \log_{10}(2) \approx 0.3010 \)). Therefore: \[ k = 2 \] ### Step 5: Substitute Values into the Freundlich Equation Now we can substitute \( k \) and the given pressure \( P = 0.8 \) atm into the Freundlich equation: \[ \frac{x}{m} = k \cdot P^{\frac{1}{n}} \] Substituting the values: \[ \frac{x}{m} = 2 \cdot (0.8)^{1} \] \[ \frac{x}{m} = 2 \cdot 0.8 \] \[ \frac{x}{m} = 1.6 \] ### Conclusion The amount of gas adsorbed per gram of charcoal under a pressure of 0.8 atm is: \[ \frac{x}{m} = 1.6 \] ### Final Answer Thus, the answer is \( 1.6 \). ---
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