JEE Mains Model Paper 7

No. of moles of acetic acid adsorbed
= (N1V1−N2V2) × n - factor
= (0.06×

50

1000

- 0.042×

50

1000

) 1
= 0.018 × 50 × 10−3 mole
= 0.018 ×

100

2

× 10−3 mole
= 0.9 × 10−3 mole
∴ wt of acetic acid adsorbed
= 0.9 × gmw × 10−3
= 0.9 × 60 milli grams
= 54 mg
∴ For 3 grams charcoal = 54 g acetic acid
∴ for 1 gram mg =

1×54

3

mg
= 18 milli grams