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

Answer:

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

Explanation:

Density = mass / volume

Volume = mass/ density

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

Answer 2

Answer:

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

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:

$D=\frac{M}{V}$

where,

D is density,

M is mass,  

V is volume.

The above formula can be rewriiten as

$V=\frac{M}{D}$

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

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

$V=318.97\frac{1 \mathrm{cm}^{3}}{19.32}$

$V=\frac{318.97}{19.32} \mathrm{cm}^{3}$

$V=16.51 \mathrm{cm}^{3} \mathrm{Au}$

Result:

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