Question

2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), and lead (d = 11.3 g/mL). If all of the samples have the same mass, which one occupies the largest volume? Why?

Answer 1

The sample of lithium occupies the largest volume.

Step-by-step explanation:

Given the densities for the four elements, we have the expression
`d=(m)/(V)` that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

`d_(lithium)=(m_(lithium))/(V_(lithium))=0.53g/mL`

`d_(gold)=(m_(gold))/(V_(gold))=19.3g/mL`

`d_(aluminum)=(m_(aluminum))/(V_(aluminum))=2.70g/mL`

`d_(lead)=(m_(lead))/(V_(lead))=11.3g/mL`

The problem says that all the samples have the same mass, so:

`m_(lithium)=m_(gold)=m_(aluminum)=m_(lead)=m`

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

`V_(lithium)=(m)/(d_(lithium))`

`V_(lithium)=(1)/(0.53(g)/(mL)) *m`

`V_(lithium)=1.88(mL)/(g)*m`

`V_(gold)=(m)/(d_(gold))`

`V_(gold)=(1)/(19.3(g)/(mL)) *m`

`V_(gold)=5.18*10^(-2)(mL)/(g)*m`

`V_(aluminum)=(m)/(d_(aluminum))`

`V_(aluminum)=(1)/(2.70(g)/(mL)) *m`

`V_(aluminum)=3.70*10^(-1)(mL)/(g)*m`

`V_(lead)=(m)/(d_(lead))`

`V_(lead)=(1)/(11.3(g)/(mL)) *m`

`V_(lead)=8.85*10^(-2)(mL)/(g)*m`

If we assume m = 1g, we find that:

`V_(lithium)=1.88mL`

`V_(gold)=5.18*10^(-2)mL`

`V_(aluminum)=3.70*10^(-1)mL`

`V_(lead)=8.85*10^(-2)mL`

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.