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

Step-by-step 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 =
`(mass)/(volume)`

Density =
`(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)}}`

=
`(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.