# I reported a volume of 8.32*mL for a mass of 6.54*g with respect to acetone, where rho_"acetone"=0.7857*g*mL^-1. Is this correct?

Sep 3, 2016

$\text{Density, } \rho$ $=$ $\text{Mass"/"Volume}$

#### Explanation:

We simply use the given equation to solve for mass or volume. We know or should know that $1 \cdot m L$ $\equiv$ $1 \cdot c {m}^{3}$.

For a $28.56 \cdot m L$ volume of acetone, $\text{Mass}$ $=$ $\text{Volume} \times \rho$ $=$ $28.56 \cdot \cancel{m L} \times 0.7857 \cdot g \cdot \cancel{m {L}^{-} 1}$ $\cong$ $22 \cdot g$

So my answer is consistent with yours (of course, we might have both made the same error, however, my answer is consistent dimensionally so this is unlikely).

For a mass of $6.54 \cdot g \text{ acetone}$,

$\text{Volume}$ $=$ $\frac{\text{Mass}}{\rho}$ $=$ $\frac{6.54 \cdot \cancel{g}}{0.7857 \cdot \cancel{g} \cdot m {L}^{-} 1}$ $=$ $\text{what you got}$, and the answer is in $\frac{1}{m {L}^{-} 1} = m L$ as required.

So, I think you are batting on a firm wicket.