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?

1 Answer
Sep 3, 2016

Answer:

#"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.