# What is the volume of 64 g of water?

$\text{Density, } {\rho}_{{H}_{2} O} = 1 \cdot g \cdot m L ,$ or $1 \cdot g \cdot c {m}^{-} 3$
Now $\rho = \text{mass"/"volume}$
And thus $\frac{\text{volume"="mass}}{\rho} = \frac{64 \cdot \cancel{g}}{1 \cdot \cancel{g} \cdot m {L}^{-} 1}$
$= \frac{64}{\frac{1}{m L}} = 64 \cdot m L$