Question

What volume would be occupied by `1.78*mol` dioxygen gas under conditions of `"STP"`?

Answer 1

Approx. `40*L`.

Explanation:

The molar volume at `"STP"` varies from syllabus to syllabus. In general, `"STP"`, `"standard temperature and pressure"` specifies `1.0*atm`, and `273*K`. It is further known that `1*mol` of an Ideal Gas occupies `22.4*L` at `"STP"`.

And thus the `"volume"` is given by the product `"number of moles"xx"molar volume"` if we assume (reasonably) that dioxygen gas behaves ideally under the given conditions.

`=1.78*cancel(mol)xx22.4*L*cancel(mol^-1)=??L`

What mass does this volume represent?

Answer 2

`39.87" L"`

Explanation:

At Standard Temperature and Pressure, `"STP" (0^(@)"C" and "1 atm")`, `"1 mole"` of any ideal gas occupies `22.4" Liters of Volume"`.

`color(white)(aaaaaaaaaaaaaaaaa)(22.4" L")/(1" mol")`

Knowing this, we can think about this for a second without involving any math or calculations.

If we know `1" mole"` of a gas occupies `22.4" Liters of volume"` at `STP`, any number of moles more than `1` would occupy more than `22.4" Liters of volume"`. How much more volume? Well.

`color(white)(aaaaaaaaaaaaaaaaa)(1.78 cancel("moles"))/(1 cancel("mole")) = 1.78`

This means, there is `1.78"X"` more number of moles. So, since we know all else is held constant (`"Temperature and Pressure"`), `"moles"` and `"Volume"` are directly proportional here so `"Volume"` will increase by the same magnitude `("Avogadro's Law")`.

`1.78 * 22.4" L" = color(blue)(39.87" L"`

Answer: 39.87 L