Calcualte the mass of
(i) 1 atom of `""^(14)C`
(ii) 1 molecule of `N_(2)`.
(iii) 1 molecule of water
(iv) 100 molecule of sucrose `(C_(12)H_(22)O_(11))`.

Explanation

(i) `6.022 xx 10^(23)` atoms of carbon = Gram atomic mass of carbon `(""^(14)C) = 14 g`
`therefore` Mass of 1 atom of `""^(14)C = 14/(6.022 xx 10^(23)) = 2.32 xx 10^(-23)`g
(ii) `6.022 xx 10^(23)` molecules of `N_(2)` = Gram molecule mass = 28g
Mass of 1 molecule of `N_(2) = 28/(6.022 xx 10^(23)) = 4.65 xx 10^(23)`g
(iii) `6.022 xx 10^(23)` molecules of sucrose = Gram molecular mass = 18 g
Mass of 1 molecule of `H_(2)O = 18/(6.022 xx 10^(23))=2.99 xx 10^(-23)` g
(iv) `6.022 xx 10^(23)` molecules of sucrose = Gram molecular mass = 342 g
Mass of 100 molecules of sucrose `=342/(6.022 xx 10^(23)) xx 100`
`=5.68 xx 10^(-20)`g