I reported a volume of 8.32*mL8.32⋅mL for a mass of 6.54*g6.54⋅g with respect to acetone, where rho_"acetone"=0.7857*g*mL^-1ρacetone=0.7857⋅g⋅mL−1. Is this correct?

1 Answer
anor277
Sep 3, 2016

"Density, "rhoDensity, ρ == "Mass"/"Volume"MassVolume

Explanation:

We simply use the given equation to solve for mass or volume. We know or should know that 1*mL1⋅mL -=≡ 1*cm^31⋅cm3.

For a 28.56*mL28.56⋅mL volume of acetone, "Mass"Mass == "Volume"xxrhoVolume×ρ == 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.

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