how many molecules are present in

a) 9g of water

b) 17g of ammonium

1 Mol of water = 18 g (2x1 for H +16 for O) = 6.023 x 1023
 9 g of water = 0.5 moles = 6.023 x 1023 / 2 = 3.011 x 1023  molecules.

Ammonia has mol wt = N (14) + 3H =17
Thus 17g of ammonia will have 6.023 x 1023 molecules.

  • 1

1. no. of atoms=Given mass/molar mass*avogadro's no.

= 9g/16*6.022*10^23

=3.38*10^23

2)no. of atoms=Given mass/molar mass*avogadro's no.

= 17/17*6.022*10^23

=1*6.022*10^23

=6.022*10^23

Thumbs up plz!!!

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