A vessel of #50*g# mass was filled with water, and the mass of the vessel and its content were #125*g#. The water was emptied, and the vessel was filled with an unknown liquid, and the mass of the vessel and it contents was #110*g#. Can you address?

#"(i) what was the density of the unknown liquid;"#
#"(ii) what was the relative density of the unknown liquid?"#

1 Answer
Jul 8, 2017

Answer:

We can infer IMMEDIATELY that the volume of the vessel is #75*mL#. Why?

Explanation:

Because we filled the vessel with #(125-50)*g=75*g# water, which has a VOLUME of #75*mL#.

When it was filled with an unknown liquid, the volume of the liquid is still #75*mL#, but the MASS of the unknown liquid was #110*g-50*g=60*g#, and thus its density, #rho#, was #"mass"/"volume"# #=# #(60*g)/(75*mL)=0.80*g*mL^-1#.

And thus #rho_"relative"=rho_"substance"/rho_"water"=(0.80*g*mL^-1)/(1.00*g*mL^-1)=0.80#, a dimensionless quantity as required.

Do you agree that this should float on water if it is immiscible with water......?

#rho_"n-pentane"=0.63*g*mL^-1#........

#rho_"n-hexane"=0.65*g*mL^-1#........

#rho_"n-heptane"=0.68*g*mL^-1#........

#rho_"n-octane"=0.70*g*mL^-1#........

#rho_"n-butyl chloride"=0.89*g*mL^-1#........

#rho_"isopropyl chloride"=0.862*g*mL^-1#........

I list the densities of some common water immiscible solvents for comparison.

On the other hand, #rho_"ethyl alcohol"=0.79*g*mL^-1#........Its density is certainly consistent (within experimental error) with that of the given liquid. Why can you argue that the unknown solvent is NOT #"ethyl alcohol"#?