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

$\rho = 1.91 \cdot k g \cdot {m}^{-} 3$
$\text{Density,}$ $\rho$ $=$ $\text{Mass"/"Volume}$. Chemists generally quote $\rho$ in terms of $g \cdot c {m}^{-} 3$, or $g \cdot {L}^{-} 1$.
And thus $\rho = \frac{42 \times {10}^{3} \cdot g}{22 \cdot {m}^{3} \times {10}^{3} \cdot L \cdot {m}^{-} 3} = 1.91 \cdot g \cdot {L}^{-} 1$