What mass of fluorine gas is associated with #0.356*L# volume, at a pressure of #45.3*mm*Hg#, and a temperature of #36.75# #""^@C#?

1 Answer
anor277
Nov 19, 2017

Approx. #30*mg#...

Explanation:

We use the Ideal Gas Law, from which #n_"molar mass"=(PV)/(RT)#.

And we also know that #1*atm# will support a column of mercury that is #760*mm# high...i.e. #1*atm-=760*mm*Hg#...

And so....

#n_"fluorine"=((45.3*mmHg)/(760*mm*Hg*atm^-1)xx0.356*L)/(0.0821*(L*atm)/(K*mol)*309.9*K)=??mol#

And given the molar quantity, we simply multiply this by the molar mass of fluorine, i.e. #38.0*g*mol^-1#. But hey, the molar mass of #F# on the Periodic Table is #19.0*g*mol^-1#. Did I make a mistake?

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