Question #682b7

1 Answer
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^@"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.