Calculate the mass of 1.51 `xx 10 ^(23)` molecule of `H_2O`.

Explanation

Mass of an atom = `("Numbre of atoms")/("Avogadro.s number") xx "Mass of" H_2O`
Mass = `(1.51 xx 10^(-23))/(6.023 xx 10^(23)) xx 18`
`= (1.51)/(6.023) xx 18`
`= (27.18)/(6.023) = 4.512 g`.
(ii) 46 g of sodium
Atomic mass of sodium = 23
Mass of sodium = 46
No. of moles = `("mass")/("atomic mass") = 46/23`
= 2 mole of sodium
(iii) `H_2O to = (1 xx 2) + (1 xx 16)`
`= 2 + 16`
`H_2O = 18g`
No of molecules = `("Avogadro Number" xx "Given mass")/("Gram molecular mass")`
`= (6.023 xx 10^(23) xx 36)/(18)`
`= 12.046 xx 10^(23)` atoms
36 g water contains `to 12.046 xx 10^(23)` atoms.