Plot of log 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, the amount of solute adsorbed per gram of adsorbent will be : (log 5=0.6990 )

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, the amount of solute adsorbed per gram of adsorbent will be : (log 5=0.6990 )

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, kis 10, the amount of solute adsorbed per gram of adsorbent will be: (log 5 = 0.6990)

Plot of log x//m against log p is a straigh line inclined at an angle of 46^(@) . When the pressure is 0.5 atom and Freundlich parameter k is 10.0 , the amount of the solute adsorbed per gram of adsorbent will be (log 5=0.6990) :

Plot of log x/m against p is a straight line inclined at an angle of 45^@ . When the pressure is 0.5 atm and Freundlich parameter, k is 10. What will be the amount of solute adsorbed per gram of adsorbent?

Plot of "log" x/m against log p is a straight line inclined at an angle of 45^@ . When the pressure is 0.5atm and k value is 10, the amount of solute adsorbed per gram of adsorbent will be