Question

Question #682b7

Answer 1

`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^@"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^@"C"` is equal to `"0.99997 g mL"^(-1)`, but since the problem didn't provide this value, you can use `"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_"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.