If it can be done with 100% efficiency, what mass of naturally occurring hydroge

Question

If it can be done with 100% efficiency, what mass of naturally occurring hydrogen gas would have to be processed to obtain a sample containing 2.60 Deuterium, 2H ^2 m H (2.0140 u m u), is sometimes used to replace the principal hydrogen isotope 1H ^1 m H in chemical studies. The percent natural abundance of deuterium is 0.015%. If it can be done with 100% efficiency, what mass of naturally occurring hydrogen gas would have to be processed to obtain a sample containing 2.60 times 1021 2H ^2 m H atoms?

Explanation / Answer

2.6 x 10^21 deuterium atom in moles = ( 2.6x10^21/6.023x10^23) = 0.004317 moles

hence we require 0.004317 moles Deuterium

since abundance = 0.015 % which means 0.015 atoms per 100 atoms Hydrogen

Hence 0.015 moles deuterium present in 100 moles H2

moles of Hydrogen required for 0.004317 moles dueterium = ( 0.004317 x 100/0.015) = 28.78

Hydrogen gas contains 2 Hydrogens

hence H2 gas moles required = 28.78/2 = 14.39

hence mass of Hydrogen required for process = moles x mol wt = 14.39 x 2 = 28.78 gm

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