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

1 Answer
Mar 29, 2017

Answer:

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

Explanation:

#"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.