For Freundlich adsorption isotherm, a plot of $$\log(x/m)$$ (y-axis) and $$\log p$$ (x-axis) gives a straight line, the intercept and slope for the line is 0.4771 and 2, respectively. The mass of gas, adsorbed per gram of adsorbent if the initial pressure is 0.04 atm is ______ $$\times 10^{-4}\,\text{g}$$.

For adsorption of a gas on a solid the Freundlich isotherm is $$\dfrac{x}{m}=k\,p^{1/n}\,.$$

Taking logarithms to the base 10, we arrive at the straight-line form

$$\log\!\left(\dfrac{x}{m}\right)=\log k+\dfrac{1}{n}\,\log p.$$

The graph of $$\log(x/m)$$ versus $$\log p$$ therefore has

$$\text{intercept}= \log k = 0.4771$$ $$\text{slope}= \dfrac{1}{n} = 2.$$

From the slope we find

$$\dfrac{1}{n}=2 \;\Longrightarrow\; n=\dfrac{1}{2}.$$

The pressure to be used is $$p = 0.04\;\text{atm}.$$ We first evaluate its logarithm:

$$0.04 = 4 \times 10^{-2}\quad\Rightarrow\quad\log(0.04)=\log 4 + \log 10^{-2}.$$

Because $$\log 4 = 0.60206$$ and $$\log 10^{-2} = -2,$$ we get

$$\log(0.04)=0.60206-2=-1.39794.$$

Substituting all known values into the linear equation,

$$\log\!\left(\dfrac{x}{m}\right)=\underbrace{0.4771}_{\log k}+ \underbrace{2}_{1/n}\times(-1.39794).$$

So

$$\log\!\left(\dfrac{x}{m}\right)=0.4771-2.79588=-2.31878.$$

To obtain $$x/m$$ we take the antilogarithm:

$$\dfrac{x}{m}=10^{-2.31878} = 10^{-2}\times 10^{-0.31878}.$$

Now $$10^{-2}=0.01$$ and $$10^{-0.31878}=\dfrac{1}{10^{0.31878}}\approx\dfrac{1}{2.083}\approx0.480,$$ hence

$$\dfrac{x}{m}=0.01 \times 0.480 = 0.00480\;\text{g\,g}^{-1}.$$

Expressing this mass in the required power of ten,

$$0.00480\;\text{g} = 4.8 \times 10^{-3}\;\text{g} = 48 \times 10^{-4}\;\text{g}.$$

So, the answer is $$48$$.