# 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?

## $\text{(i) what was the density of the unknown liquid;}$ $\text{(ii) what was the relative density of the unknown liquid?}$

Jul 8, 2017

We can infer IMMEDIATELY that the volume of the vessel is $75 \cdot m L$. Why?

#### Explanation:

Because we filled the vessel with $\left(125 - 50\right) \cdot g = 75 \cdot g$ water, which has a VOLUME of $75 \cdot m L$.

When it was filled with an unknown liquid, the volume of the liquid is still $75 \cdot m L$, but the MASS of the unknown liquid was $110 \cdot g - 50 \cdot g = 60 \cdot g$, and thus its density, $\rho$, was $\text{mass"/"volume}$ $=$ $\frac{60 \cdot g}{75 \cdot m L} = 0.80 \cdot g \cdot m {L}^{-} 1$.

And thus ${\rho}_{\text{relative"=rho_"substance"/rho_"water}} = \frac{0.80 \cdot g \cdot m {L}^{-} 1}{1.00 \cdot g \cdot m {L}^{-} 1} = 0.80$, a dimensionless quantity as required.

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

${\rho}_{\text{n-pentane}} = 0.63 \cdot g \cdot m {L}^{-} 1$........

${\rho}_{\text{n-hexane}} = 0.65 \cdot g \cdot m {L}^{-} 1$........

${\rho}_{\text{n-heptane}} = 0.68 \cdot g \cdot m {L}^{-} 1$........

${\rho}_{\text{n-octane}} = 0.70 \cdot g \cdot m {L}^{-} 1$........

${\rho}_{\text{n-butyl chloride}} = 0.89 \cdot g \cdot m {L}^{-} 1$........

${\rho}_{\text{isopropyl chloride}} = 0.862 \cdot g \cdot m {L}^{-} 1$........

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

On the other hand, ${\rho}_{\text{ethyl alcohol}} = 0.79 \cdot g \cdot m {L}^{-} 1$........Its density is certainly consistent (within experimental error) with that of the given liquid. Why can you argue that the unknown solvent is NOT $\text{ethyl alcohol}$?