Hi Sabrina G.,
Strategy: From a volume and a density we can find mass. And from mass we can find moles and molecules.
We are given the total volume of the oceans in part (d) as 1.35 billion km3 or, VT = 1.35 x 109 km3.
Part (c) states 96.5% of VT (T for Total) is water and 3.5% is salts. Lets call this value Vw (w for water).
95.5%*VT = 95.5%*1.35 x 109 km3 = 1.29 x 109 km3 = Vw.
Next we have two different densities for the two different temperatures. The top 10% of water has a different density than the bottom 90%. So we will need to know the volume of the top and the bottom.
10%*Vw = 10%*1.29 x 109 km3 = 1.29 x 108 km3 = V10
90%*Vw = 90%*1.29 x 109 km3 = 1.16 x 109 km3 = V90
We now know the volume of surface (10%) water and the volume of deep (90%) water. We can use the densities to find the grams of each, then we can use Avogadro's number to find the molecules of water. Lets first change the densities from kg/m3 to g/km3, since we have km3 in our volumes and grams per mole.
1027 kg/m3 *(1000 m/1 km)3 *(1000 g/ 1 kg) = 1.027 x 1015 g/km3.
1028 kg/m3 = 1.028 x 1015 g/km3. (since this density is 1 more than the other)
Now lets find the grams of water for each volume.
For V10, 1.29 x 108 km3 *(1.027 x 1015 g/km3) = 1.32 x 1023 g.
For V90, 1.16 x 109 km3 *(1.028 x 1015 g/km3) = 1.19 x 1024 g.
Now lets find the molecules for each mass. H2O = 18.02 g/mol
For V10 mass, 1.32 x 1023 g *(1 mol/18.02 g)*(6.022 x 1023 molecules/mol) = 4.41 x 1045 H2O molecules.
For V90 mass, 1.16 x 1024 g *(1 mol/18.02 g)*(6.022 x 1023 molecules/mol) = 3.88 x 1046 H2O molecules.
Now we have the molecules for each. Lets add them for the total ocean H2O molecules.
4.41 x 1045 + 3.88 x 1046 = 4.32 x 1046 H2O molecules in the ocean.
Be sure to check the math. If you want total atoms, remember 1 H2O molecules = 3 atoms.
I hope this helps, Joe.
