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 g.
(log 3 = 0.4771)

According to Freundlich adsorption isotherm, in the median range of pressure <br><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;${x \over m} \propto {P^{{1 \over n}}}$ <br><br>$\Rightarrow$&nbsp;&nbsp;${x \over m}$ = kP$^{{1 \over n}}$ <br><br>taking log both sides, we get, <br><br>log${x \over m}$ = logk + ${1 \over n}$ logP <br><br>Here in graph between log${x \over m}$ and logP, slope is ${1 \over n}$ and intercepts = log k. <br><br>$\therefore$ ${1 \over n}$ = 2 $\Rightarrow$ n = ${1 \over 2}$ <br><br>and log k = 0.4771 = log 3 <br><br>$\Rightarrow$ k = 3 <br><br>$\therefore$ ${x \over m}$ = 3P$^{{2}}$ <br><br>So, mass of gas adsorbed per gram of adsorbent <br><br>= 3 $\times$ (0.04)² <br><br>= 48 $\times$ 10^–4