Plot of log against log P is a str...

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 )

1 g

2 g

3 g

5 g

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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) :

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:

Graph between log x/m and log P is a straight line at angle of 45^(@) with intercept 0.4771 on y-axis. Calculate the amount of gas adsorbed in gram per gram of adsorbent when pressure is 3 atm.

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)

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