Step 1

Answer to part (a):

In this part, we have been asked to calculate the volume of the given material for which we will use the below relation :

\(\displaystyle{V}{o}{l}{u}{m}{e}={\frac{{{M}{a}{s}{s}}}{{{D}{e}{n}{s}{i}{t}{y}}}}\) (1)

Given, \(\displaystyle{m}{a}{s}{s}={0.715}{k}{g}\) and \(\displaystyle{d}{e}{n}{s}{i}{t}{y}={19.3}\frac{{g}}{{c}}{m}^{{{3}}}\)

In order to calculate the density in \(\displaystyle{c}{m}^{{{3}}}\), we will first convert the mass in gram, for which we will use the below relation :

\(\displaystyle{0.715}{k}{g}\times{\frac{{{1000}{g}}}{{{1}{k}{g}}}}={715}{g}\) (2)

Step 2

Answer to part (a): Given, \(\displaystyle{m}{a}{s}{s}={715}{g}\) and \(\displaystyle{d}{e}{n}{s}{i}{t}{y}={19.3}\frac{{g}}{{c}}{m}^{{{3}}}\)

Putting above given values in equation (1) and rounding off the volume to 3 significant digits, we get :

\(\displaystyle{V}{o}{l}{u}{m}{e}={\frac{{{715}{g}}}{{{19.3}\frac{{g}}{{c}}{m}^{{{3}}}}}}={37.0}{c}{m}^{{{3}}}\) (3)

Answer to part (b):

In this problem, we have been asked to calculate the length of each edge of gold (cube), for which we will use the below relation :

\(\text{Volume of gold=(length of each edge of gold)}^{3}\) (4)

As calculated in the above subpart that the volume of the \(\displaystyle{g}{o}{l}{d}={37.0}{c}{m}^{{{3}}}\), putting this to equation (4), we get :

\(length=(37 cm{^3})^{1/3}=3.33 cm\) (5)