1g of charcoal having surface area 3.02*10?m?is mixed in 100 ml of 0.5 M acetic acid solution; after few times concentration of solution becomes 0.49 M what will be surface area occupied one molecule of acetic acid ?
moles=molarity x volume (L)
= (0.5 - 0.49) x 100 x 10-3
= 10-3 moles
= 10-3 x NA
= 6.02 x 1020 molecules
= (0.5 - 0.49) x 100 x 10-3
= 10-3 moles
= 10-3 x NA
= 6.02 x 1020 molecules