Question

A sample of gasoline has a mass of 595.0 lb and a volume of 14.0 ft^3. What is its specific gravity?

Answer 1

Step-by-step explanation:

The given data is as follows.

mass of gasoline = 595.0 lb

volume of gasoline =
`14.0 ft^(3)`

As it is known that density is the amount of mass of a substance divided by its volume.

So, density of gasoline will be as follows.

Density =
`(mass)/(volume)`

=
`(595.0 lb)/(14.0 ft^(3))`

= 42.5
`lb/ft^(3)`

As, 1 lb = 0.4536 kg and 1 ft = 0.3048 m. Now, putting these values into the above equation (1) as follows.

Density = 42.5
`lb/ft^(3)`

=
`42.5 * (0.4536 kg)/((0.3048 m)^(3))`

= 680.792
`kg/m^(3)`

= 680.8
`kg/m^(3)` (approx)

Density of water is 1000
`kg/m^(3)`. To measure specific gravity of gasoline, the formula will be as follows.

specific gravity of gasoline =
`\frac{\text{density of gasoline}}{\text{density of water}}`

=
`(680.8 kg/m^(3))/(1000 kg/m^(3))`

= 0.681

Thus, we can conclude that the specific gravity of gasoline is 0.681.