Find the number of moles and molecules present in `7.1 g of Cl_(2). (At Wt. Cl = 35.5)`

Explanation

No. of moles `= ("Mass of substance")/("Mass of one mole")= (7.1)/(71) = 0.1` moles of `Cl _(2)`
71 g of chlorine has `6. 0 22 cc 10 ^(23)` molecules
`7.1 g ` of chlorine will have `= (6. 0 22 xx 10 ^(23))/(71) xx 7.1 = 6. 0 22 cc 10 ^(22)` molecules