A #42kg# mass of substance occupies a volume of #22m^3#...what is #rho_"material"# in #g*L^-1#?

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Answer 1 · 2017-03-29T16:05:27

#rho=1.91*kg*m^-3#

#"Density,"# #rho# #=# #"Mass"/"Volume"#. Chemists generally quote #rho# in terms of #g*cm^-3#, or #g*L^-1#.

And thus #rho=(42xx10^3*g)/(22*m^3xx10^3*L*m^-3)=1.91*g*L^-1#

I think the gas (and how did I know it was a gas?) is one of the heavier ones, possibly carbon dioxide or argon.

A #42*kg# mass of substance occupies a volume of #22*m^3#...what is #rho_"material"# in #g*L^-1#?