The natural abundances of the two stable isotopes of hydrogen (hydrogen and deuterium) are \({ }_{1}^{1} \mathrm{H}\) : 99.985 percent and \({ }_{1}^{2} \mathrm{H}: 0.015\) percent. Assume that water exists as either \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{D}_{2} \mathrm{O} .\) Calculate the number of \(\mathrm{D}_{2} \mathrm{O}\) molecules in exactly \(400 \mathrm{~mL}\) of water. (Density \(=1.00 \mathrm{~g} / \mathrm{mL} .)\)

Short Answer

Expert verified
The number of \(\mathrm{D}_{2}\mathrm{O}\) molecules in 400 mL of water is approximately \(9.033 \times 10^{22}\).

Step by step solution

01

Calculate mass of water

First, find the mass of the water. We know, density = mass/volume. so, mass = density * volume = \(1.00 \, \mathrm{g/mL} * 400 \, \mathrm{mL} = 400 \, \mathrm{g}\).
02

Calculate mass of Deuterium

Next, find the mass of Deuterium. Assume the water exists only as \( \mathrm{H}_{2}\mathrm{O} \) and \( \mathrm{D}_{2}\mathrm{O} \). As such, 0.015 percent of the hydrogen atoms are Deuterium. So, the mass of deuterium = \(0.00015 * 400 \, \mathrm{g} = 0.06 \, \mathrm{g}\).
03

Compute moles of Deuterium

Then, convert the mass of Deuterium to moles. The molar mass of Deuterium is 2.00 g/mol. So, moles of Deuterium = \(0.06 \, \mathrm{g} / 2.00 \, \mathrm{g/mol} = 0.03 \, \mathrm{mol}\).
04

Calculate molecules of \( \mathrm{D}_{2}\mathrm{O} \)

Finally, calculate the number of \( \mathrm{D}_{2}\mathrm{O} \) molecules. As each \( \mathrm{D}_{2}\mathrm{O} \) molecule contains two Deuterium atoms, the moles of \( \mathrm{D}_{2}\mathrm{O} \) is half the moles of Deuterium. So, moles of \( \mathrm{D}_{2}\mathrm{O} \) = \(0.03 \, \mathrm{mol} / 2 = 0.015 \, \mathrm{mol}\). Then, to get the number of molecules, multiply by Avogadro's number \(6.022 \times 10^{23} \, \mathrm{mol}^{-1}\). Therefore, the number of \( \mathrm{D}_{2}\mathrm{O} \) molecules = \(0.015 \, \mathrm{mol} * 6.022 \times 10^{23} \, \mathrm{mol}^{-1} = 9.033 \times 10^{22}\).

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