1.

Graph between log x/ m and log P is a straight line inclined at an angle theta – 45^@. When pressure of 0.5 atm and log k = 0.699, the amount of solute adsorbed per g of adsorbent will be:

Answer»

1g/g adsorbent
1.5g/g adsorbent
2.5g/g adsorbent
0.25g/g adsorbent

Solution :`x/m=k.p^(1//n) since logk=0.699 HENCE k=5, Hence k=5`
Slope =`1/5= tan45^@=1 THUS, x/m=5 TIMES 0.5=2.5g//g` adsorbent


Discussion

No Comment Found

Related InterviewSolutions