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3 years ago

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

Chemistry

2 answers:

zmey [24]3 years ago

8 0

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.

Colt1911 [192]3 years ago

6 0

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

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