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

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 = $\frac{mass}{volume}$

                              = $\frac{595.0 lb}{14.0 ft^{3}}$

                              = 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 $lb/ft^{3}$

                                     = $42.5 \times \frac{0.4536 kg}{(0.3048 m)^{3}}$

                                     = 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 = $\frac{\text{density of gasoline}}{\text{density of water}}$    

                                         = $\frac{680.8 kg/m^{3}}{1000 kg/m^{3}}$

                                         = 0.681

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