# Question 682b7

Feb 15, 2018

$0.64$

#### Explanation:

The specific gravity of a substance is usually taken to mean the ratio that exists between the density of the substance and the density of water at ${4}^{\circ} \text{C}$, the temperature at which the density of water is at its maximum value.

"SG"_ "substance" = rho_"substance"/rho_ ("water at 4"^@"C")

Notice that the specific gravity is a dimensionless quantity because you're dividing two densities.

Now, the density of water at ${4}^{\circ} \text{C}$ is equal to ${\text{0.99997 g mL}}^{- 1}$, but since the problem didn't provide this value, you can use ${\text{1 g mL}}^{- 1}$ as the density of water.

So, calculate the density of the substance by dividing the mass of the sample by the volume it occupies.

${\rho}_{\text{substance" = "1.6 g"/"2.5 mL" = "0.64 g mL}}^{- 1}$

The *specific gravity of the substance will thus be

"SG" = (0.64 color(red)(cancel(color(black)("g mL"^(-1)))))/(1 color(red)(cancel(color(black)("g mL"^(-1))))) = color(darkgreen)(ul(color(black)(0.64)))#

The answer is rounded to two sig figs.