The graph between $ \log \dfrac{X}{m}\;{\text{vs }}\log P $ is straight line inclined at an angle $ {45^0} $ with intercept $ 0.30 $ . What will be the rate of adsorption at pressure of $ 0.4atm $ ?
(A) $ 0.4 $
(B) $ 0.6 $
(C) $ 0.8 $
(D) $ 0.9 $

Hint: The rate of adsorption at a particular pressure is the amount of a gas adsorbed per unit mass of solid adsorbent. The relationship is commonly known as the Freundlich adsorption isotherm. We will use the relationship to calculate the rate of adsorption.

Formula used:
Freundlich adsorption isotherm, $ \dfrac{x}{m} = K{P^{\dfrac{1}{n}}} $ .
Where, $ x $ is the mass of gas adsorbed, $ m $ is the mass of adsorbent, $ \dfrac{x}{m} $ is the rate of adsorption, $ P $ is the pressure and $ K $ is any constant.

Complete step by step answer:
Now we are aware about the Freundlich adsorption isotherm. The Freundlich adsorption isotherm gives the relationship between the magnitude or the rate of adsorption and pressure which van be expressed mathematically as, $ \dfrac{x}{m} = K{P^{\dfrac{1}{n}}} $ . If we take logarithm on both sides we get the equation as, $ \log \dfrac{x}{m}\; = \log K + \dfrac{1}{n}\log P $ . Here we have used the basic properties of logarithm, the product and the power rule.
Now we can relate the above equation which also represents the equation of straight line. So, now in the question we have given that the straight line inclined at an angle $ {45^0} $ which is equal to the slope with intercept $ 0.30 $ . So in the equation, $ \log \dfrac{x}{m}\; = \log K + \dfrac{1}{n}\log P $ the slope is $ \dfrac{1}{n} = \tan {45^0} = 1,n = 1 $ and the intercept is $ \log K = 0.30,K = 2 $ . Now we have all the unknowns so we can easily calculate the rate of adsorption at the given quantities as $ P = 0.4,K = 2,n = 1 $ . We will substitute in the relationship $ \dfrac{x}{m} = K{P^{\dfrac{1}{n}}} $ we get,
 $ \dfrac{x}{m} = K{P^{\dfrac{1}{n}}} = (2){(0.4)^1} = 0.8 $
Hence, the rate of adsorption at pressure of $ 0.4atm $ is $ 0.8 $ .
Therefore, the correct option is (C).

Note:
There is one limitation in the relationship of the Freundlich adsorption isotherm. The isotherm fails in case of high pressure and high concentration of adsorbate. The relationship is Freundlich adsorption isotherm only when the range of $ \dfrac{1}{n} $ is between $ 0 $ and $ 1 $ .