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*kg42*kg mass of substance occupies a volume of 22*m^322*m^3...what is rho_"material"rho_"material" in g*L^-1g*L^-1?