Graph between `log x//m` and `log p` is a straight line inclined at an angle of `45^@` . When pressure is `0.5 atm` and `1n k = 0.693`, the amount of solute adsorbed per gram of adsorbent will be:

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 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) :

When a graph is plotted between log x/m and log p, it is straight line with an angle 45^(@) and intercept 0.3010 on y - axis. If initial pressure is 0.3 atm, what will be the amount of gas adsorbed per gram of adsorbent : (Report your answer after multiplying by 10)

When a graph is plotted between log x//m and log p, it is straight line with an angle 45^(@) and intercept 0.3010 on y- axis . If initial pressure is 0.3 atm, what wil be the amount of gas adsorbed per gm of adsorbent :