Question

Question #d3948

Answer 1

`"0.002 g mL"^(-1)`

Explanation:

The density of a substance, `rho`, tells you the mass of exactly one unit of volume of said substance.

In your case, the sample of carbon dioxide has a mass of `"100 mL"`, so you can say that one unit of volume will be `"1 mL"`. This means that your goal here is to figure out the mass of `"1 mL"` of carbon dioxide.

Now, you know that the mass of `"100 mL"` of carbon dioxide is equal to `"0.196 g"`.

You can find the density of carbon dioxide by dividing the mass of the sample by the total volume it occupies

`rho = "0.196 g"/"100 mL"`

since this will get you the mass of `"1 mL"` of carbon dioxide

`rho = "0.196 g"/(100 * "1 mL") = "0.196 g"/100 * 1/"1 mL"`

`= "0.00196 g" * 1/"1 mL" = "0.00196 g"/"1 mL"`

`= color(darkgreen)(ul(color(black)("0.002 g mL"^(-1))))`

The answer must be rounded to one significant figure, the number of sig figs you have for the volume of the sample.

You can thus say that `"1 mL"` of carbon dioxide has a mass of `"0.002 g"`.

SIDE NOTE We usually express the density of a gas in grams per liter, `"g L"^(-1)`.

In your case, the density of carbon dioxide in grams per liter is equal to

`0.002 color(white)(.)"g"/(1color(red)(cancel(color(black)("mL")))) * (10^3color(white)(.)color(red)(cancel(color(black)("mL"))))/"1 L" = "2 g L"^(-1)`

This is very close to the density of carbon dioxide at STP conditions

https://en.wikipedia.org/wiki/Carbon_dioxide