# Question #46628

##### 1 Answer

#### Explanation:

The first thing that you should know here is that

#color(blue)(ul(color(black)("1 mL" = "1 cm"^3)))#

This means that you can express the volume of the can as

#"355 mL" = "355 cm"^3#

Now, the **density** of platinum tells you the mass of exactly **every**

You can thus use the density of platinum as a *conversion factor* to determine the *mass* of platinum present in the sample.

#355 color(red)(cancel(color(black)("cm"^3))) * overbrace("21.4 g"/(1color(red)(cancel(color(black)("cm"^3)))))^(color(blue)("the density of platinum")) = "7,597 g"#

Finally, you know that you can get **per gram** of platinum, which means that the sample will have a value of

#"7,597" color(red)(cancel(color(black)("g"))) * ($70.00)/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("$532,000")))#

The answer is rounded to three **sig figs**, the number of sig figs you have for the volume of the can.