Question

A sample of gold has a mass of 645.0 lb and a volume of 0.5348 ft. What is its specific gravity? Number SG= What are the units of specific gravity? Ob/t kg/m2 g/cm none

Answer 1

Explanation:

The given data is as follows.

        Mass of gold = 645 lb,       Volume = 0.5348 ft3

         Density of water = 62.4 lbs/ft3

It is known that specific gravity is defined as density of substance divided by the density of standard fluid.

Mathematically,       Specific gravity = $\frac{\text{density of gold}}{\text{density of water}}$

               Specific gravity = $\frac{\text{density of gold}}{62.4 (lbs.ft^{-3})}$

Now, calculate the density of gold then from density we will calculate specific gravity as follows

Since,             Density = $\frac{mass}{volume}$

                        Density = $\frac{645 lbs}{0.5348 ft^{3}}$

                                    = 1206.06 $lbs/ft^{3}$

As,        Specific gravity = $\frac{\text{density of gold}}{\text{density of water (standard fluid)}}$

                                      = $\frac{1206.06 (lbs/ft^{3})}{62.4 (lbs/ft^{3})}$

                                      = 19.32

Therefore, the value of specific gravity is 19.32.

Specific gravity has no units as it is density divided by density. Hence, all the units get canceled out.