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The weight of 350 mL of a diatomic gas a...

The weight of 350 mL of a diatomic gas at `0^(@)C` temperature and 2 atm pressure is 1g. Calculate the weight of its one atom.

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According to the given condition, volume of diatomic gas at `0^(@)C` and 2 atm pressure=350 mL
Let, the volume of the gas at `0^(@)C` and 1 atm pressure be V mL
Now, according to Boyle's law `P_(1)V_(1)=P_(2)V_(2)`
or, `2xx350=1xxV" "therefore V=700mL`
i.e., mass of 700 mL of the gas at `0^(@)C` and 1 atm pressure =1g
`therefore`Mass of 22400 mL of the gas at `0^(@)C` and 1 atm pressure
`=(22400)/(700)xx1=32g`
Molar volume of all gases at STP=22400 mL.
Hence, the gram-molecular mass of the given gas=32g.
So, the mass of `6.022xx10^(23)` molecules of the given gas=32g.
Therefore, the mass of `2xx0.622xx10^(23)` atoms of the gas=32g. [since the gas is diatomic]
Thus, the mass of 1 atom of the given gas
`=(32)/(2xx6.022xx10^(23))=2.656xx10^(-23)g`.
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