Question

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?

Answer 1

`"Density, "rho` `=` `"Mass"/"Volume"`

Explanation:

We simply use the given equation to solve for mass or volume. We know or should know that `1*mL` `-=` `1*cm^3`.

For a `28.56*mL` volume of acetone, `"Mass"` `=` `"Volume"xxrho` `=` `28.56*cancel(mL)xx0.7857*g*cancel(mL^-1)` `~=` `22*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*g" acetone"`,

`"Volume"` `=` `"Mass"/rho` `=` `(6.54*cancelg)/(0.7857*cancelg*mL^-1)` `=` `"what you got"`, and the answer is in `1/(mL^-1)=mL` as required.

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