# Answer the following without doing detailed calculations. Which has the greater volume, a 50.0g block of copper or a 50.0g block of gold? I don't understand their explanation. From table they give densities of gold and copper. Which I understand. But I don't understand the second part. A block of copper thus has to be larger in volume than a block of gold of the same mass. A 50.0 g block of copper has a greater volume that 50.0 block of gold.

OK... without a bunch of math. Think of it like this:

Which weighs more, a ton of feathers or a ton of bricks? Right, they both weigh the same (just like the gold and copper in your example). BUT, which one has a bigger volume? Or, which one would make...

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OK... without a bunch of math. Think of it like this:

Which weighs more, a ton of feathers or a ton of bricks? Right, they both weigh the same (just like the gold and copper in your example). BUT, which one has a bigger volume? Or, which one would make a bigger pile (take up more space)?

Obviously, one feather is much lighter than one brick. This is because bricks are more dense than feathers (they have more particles crammed into the same amount of space). So it's going to take a really big pile of feathers (greater volume) to equal one ton.

Same thing in your example... copper is not as dense as gold, so you are going to need a lot more of it to make 50.0g than you would 50.0 g of gold.

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To determine which has a greater volume, we need to convert the given mass to volume.

To do so, let's use the density of copper and gold.

The density of copper is 8.96 `g/(cm^3)` . This means that for every `1 cm^3` of

copper, it weighs 8.96 g ( `1cm^3=8.96g` ).

So, the volume of 50g copper is:

`V_(copper)``=` `50 g` `xx` `(1 cm^3)/(8.96g)` `=`  `5.58 cm^3`

And the density of gold is 19.32 `g/(cm)^3` . This indicates that `1cm^3=19.32g` .

Hence, the equivalent volume of 50g gold is:

`V_(gold)` `=` `50 g` `xx` `(1cm^3)/(19.32g)` `=` `2.58cm^3`

Therefore, the 50g block of copper has greater volume than the 50g block of gold.

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