Question

Riley discovered pure gold has a density of 19.32 g/cm3.What would the volume for a piece of gold be if it had a mass of 318.97 g?

Answer 1

The volume of a gold piece is
`V=16.51 \mathrm{cm}^(3)`

Step-by-step explanation:

Given:

Density of pure gold,
`D=19.32 \frac{g}{\mathrm{cm}^(3)  }`

Mass of the gold piece,
`\mathrm{M}=318.97`

To Find:

The volume for a piece of gold whose mass is
`318.97 g`

Solution:

The formula for density is:

`D=(M)/(V)`

where,

D is density,

M is mass,

V is volume.

The above formula can be rewriiten as

`V=(M)/(D)`

Substituting the values ,we get

`V=318.97 \mathrm{g} \frac{\mathrm{Au}}{19.32 \frac{\mathrm{g}}{\mathrm{cm}^(3)} \mathrm{Au}}`

Taking multiplicative inverse of density , we have

`V=318.97 \text { of }Au\frac{1 \mathrm{cm}^(3)\mathrm{Au}}{19.32 \mathrm{of}\mathrm{Au}}`

`V=318.97\frac{1 \mathrm{cm}^(3)}{19.32}`

`V=(318.97)/(19.32) \mathrm{cm}^(3)`

`V=16.51 \mathrm{cm}^(3) \mathrm{Au}`

Result:

Thus the volume of gold piece weighing 318.97 g is
`V=16.51 \mathrm{cm}^(3)`

Answer 2

The volume for a piece of gold that has the mass of 318,67 g is 16,50cm3

Step-by-step explanation:

Density = mass / volume

Volume = mass/ density

Volume = 318,97 g / 19,32 g/cm3 = 16,50cm3