help with section 4, question e. 4. Lake Tahoe has an approximate volume of 36 c
ID: 540272 • Letter: H
Question
help with section 4, question e.
Explanation / Answer
Ans > Volume of the lake is 36 cubic miles = 36 x 4.168 cubic kilometers = 150.048 x 109 m3 [ 1 mile3 = 4.168km3] = 1.5 x 1011 m3 Density of the water present is 1g/ml = 1000 kg/m3 Total mass of water present in the lake is 1.5 x 1014 kg Now suppose that a% of the hydrogen atom are deuterium then the molar mass of D2O would be (2 x 2.01410)+16 = 20.0282g/mole. Now the remaining (100-a)% would be hydrogen atoms and the molar mass of H2O would be (2 x 1.00783) + 16 = 18.01566g/mole.
But the average molar mass of hydrogen present is 1.00794 g/mole. Then the avergae molar mass of water molecules present would be (2 x 1.00794)+16 = 18.01588g/mole = 18.01558 x 10-03kg/mole Total number of molecules present is 1.5 x 1014kg/(18.01558 x 10-03) = 8.326 x 1015moles
Now, if a% of molecules are D2O then there is (a/100) fractions of D2O and simmlilarly there is (100-a)/100) fractions of H2O. Therefore, (a/100)MD20 + {(100-a)/100}MH20 = average mass of the water present => (a/100)20.0282 + {(100-a)/100}18.01556 = 18.01558 => 20.0282a + 1801.556 - 18.01556a = 1801.558 => 2.01264a = 0.002 => a = 0.0009918% Deuterium is present Total number of moles of deuterium present is (0.0009918/100) x (8.326 x 1015) = 8.25 x 1010moles of deuterium is present in the Lake Tahoe
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