Answer:
[tex]m_{Deu}=2666.7g[/tex]
Explanation:
Hello,
In this case, we can apply the following mole-mass-density relationship in order to compute the required grams of deuterium, considering that it is the 0.0150% (molar basis) of natural hydrogen (Hâ‚‚):
[tex]m_{Deu}=80000LH_2O*\frac{1000gH_2O}{1LH_2O}*\frac{1molH_2O}{18gH_2O} *\frac{1molH_2}{1molH_2O} *\frac{2gH_2}{1molH_2} *\frac{2*0.0150g\ Deu}{100gH_2} \\\\m_{Deu}=2666.7g[/tex]
Best regards.
