Plot of log x/m 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 10.6, the amount of the solute adsorbed per gram ofadsorbent will be ( log 5 =0.0990)

Plot of log x/m against log p is a straight line inclined at an angle

1.	Plot of log x/m 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 10.6, the amount of the solute adsorbed per gram ofadsorbent will be ( log 5 =0.0990)
Answer» 1 g 2g 3g 5g Solution :According to Freunlich EQUATION , `x//m=kp^(1//n)` or `LOG(x)/(m)log k+(1)/(n) logp` `therefore` plot of log x/m vs log p is a straight LINE with slope=1/n and intercept = log k. Thus, `(1)/(n)= tan theta = tan 45^(@)=1 " or " n=1` At p=0.5 atm and k=10 x/m =10(0.52)=5 When m=1g `therefore` x (i.e., amount adsorbed PER gram )=5 g