Question

How many grams of water are in 4.589 x 10^23 molecules of water (H2O)?

Answer 1

Answer is: 13.7 grams of water.

N(H₂O) = 4.589·10²³; number of water molecules.

n(H₂O) = N(H₂O) ÷ Na(Avogadro constant).

n(H₂O) = 4.589·10²³ ÷ 6.022·10²³ 1/mol.

n(H₂O) = 0.76 mol; amount of substance.

M(H₂O) = 2 · Ar(H) + Ar(O) · g/mol.

M(H₂O) = 2 · 1 + 16 · g/mol.

M(H₂O) = 18 g/mol; molar mass of water.

m(H₂O) = n(H₂O) · M(H₂O).

m(H₂O) = 0.76 mol · 18 g/mol.

m(H₂O) = 13.7 g.

Answer 2

1 mole    6.023  x10^23  from  avogadro laws  which  state  that  equal  volume  of  a  gases at  the  same  temperature  and  pressure  contain  equal   number  of  moles.
what  about  4.589 x  10^23
=4.589  x10^23/6.023 x10^23=0.762moles
mass=molar  mass  x  number  of  moles
18  x 0.762=13.716grams