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dangina [55]
3 years ago
7

Deuterium, 2h (2.0140 u), is sometimes used to replace the principal hydrogen isotope 1h in chemical studies. the percent natura

l abundance of deuterium is 0.015%. part a 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 2.50Ã1021 2h atoms?
Chemistry
1 answer:
Pavlova-9 [17]3 years ago
4 0

Deuterium is isotope of hydrogen with percent natural abundance 0.015%. Thus, if 100 gram of naturally occurring hydrogen gas is taken it will have 0.015 g of deuterium.

Or,

0.015g  ^{2}H\rightarrow 100 g ^{1}H

or,

1 g^{2}H\rightarrow 6666.67 g ^{1}H

The number of deuterium atoms are given 2.50\times 10^{21}.

Since, 1 mol is equal to 6.023\times 10^{23} atoms thus,

1 atom\rightarrow \frac{1}{6.023\times 10^{23}}mol=1.66\times 10^{-24}mol

Thus,  2.50\times 10^{21} will be equal to 2.50\times 10^{21}\times 1.66\times 10^{-24}=0.00415 mol

Molar mass of deuterium is 2.0140 u or 2.0140 g/mol thus, mass can be calculated as:

m=n\times M=0.00415 mol\times 2.0140 g/mol=0.00836 g

thus, mass of ^{2}H is 0.00836 g

Since, 1 g^{2}H\rightarrow 6666.67 g ^{1}H

Thus, 0.00836 g ^{2}H\rightarrow 0.00836\times 6666.67=55.73 g ^{1}H

Therefore, mass of naturally occurring hydrogen gas that has to be processed is 55.73 g.

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