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

Explanation

`"Mass of an atom " = ("Number 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 soidum = 23
Mass of sodium =46
No. of moles `=("mass ")/("atomic mass ") = 46/23`
2 mole of sodium
(iii) `H_(2)O rarr = (1 xx 2) + (1 xx 16)`
= 2 + 16
`H_(2)O = 18 g `
NO of molecule `= ("Avorgadro Number" xx " Given mass ")/("Gram molecular mass ")`
`= (6.023 xx 10^23 xx 36)/(18)`
`= 12.046 xx 10^23` atoms
36g water contains `rarr 12.046 xx 10^23` atoms