Question

What volume is occupied by a `64*g` mass of oxygen gas confined in a piston under conditions of `"STP"`?

Answer 1

Approx. `44*L`..........

Explanation:

The molar volume of `"STP"` is usually quoted at `22.4*L*mol^-1` (these definitions vary according to the syllabus you follow....

For the molar quantity of this mass of dioxygen gas we have....

`(64.0*g)/(32.0*g*mol^-1)=2*mol`....

And so to find the volume occupied by the molar quantity, we take the product....

`"Moles"xx"molar mass"` `=` `2*molxx22.4*L*mol^-1=??*L`

Note that you must simply know that ALL elemental gases SAVE the NOBLE GASES, are bimolecular, i.e. dinitrogen, dioxygen, difluorine, dichlorine....

Answer 2

Use the molar volume (`(22.4 "L")/("mol"`). This value can only be used at `"STP"` conditions `(0^@" C" " or " 273 " K", and 1 " atm")`.

Explanation:

We can use dimensional analysis to convert from:

`color(red)("g") " O"_2 -> color(red)(" moles ") "of" " O"_2 -> color(red)("L ")"of" " O"_2 " at STP"`

First, let's calculate the molar mass of `"O"_2` using the Periodic Table:

`"O"_2 = 2(16.00) = 32.00 "g"/"mol"`

Now, we are ready to begin:

`64.0" g" " O"_2 ((1" mol" " O"_2)/(32.00" g" " O"_2))((22.4" L")/(1" mol")) = 44.80 " L O"_2`

Your final answer is therefore `44.8 " L O"_(2("g"))` with significant figures and proper units.

I hope that helps!

Answer 3

`45 dm^3` rounded.

Explanation:

Molar mass of `O_2` molecule is 31.9

Number of moles = mass in grams/ molar mass.

`64.0/31.9= 2.01` moles.

1 mole of of `O_2` at S.T.P = `22.4 dm^3`

So:

`2.01 xx22.4= 45.024 dm^3`

Answer 4

48 `dm^3` OR 44.8 `dm^3`

Explanation:

At GCSE level, I was taught that one mole of gas occupies a volume of 24 `dm^3` (or 24 litres). However when I searched online and learned that the accurate volume 1 mole of gas occupies is 22.4 `dm^3` (or 24 litres).

I will show you the calculation for both.

First we need to calculate how many moles of `O_2` are in 64.0g using the equation: `(mass)/(mr)`=moles

mass=64.0g
Mr( `O_2`)= `16*2`=32

moles= `64.0/32=2`

If 1 mole of `O_2` = `24dm^3`
2 moles of `O_2` = `48dm^3`

If 1 mole of `O_2` = `22.4dm^3`
2 moles of `O_2` = `44.8dm^3`