Question

What volume would a `10*g` mass of carbon monoxide enclosed in a piston occupy at `1*atm`, and `0` `""^@C`?

Answer 1

Well, one mole of ideal gas occupies 22.4 L.

Explanation:

So, we have `(10*g)/(28.01*g*mol)~=1/3*"mole"`.

And thus if we assume carbon monoxide behaves ideally, its volume (if enclosed in a piston expanding against atmospheric pressure) would be approx. `7.4*L`.

Answer 2

8.00 Liter

Explanation:

It can be further explained as :-

idle gas equation is

`PV=nRT`

Here ,

P = 1 atm
V = To be find out
R = 0.08206 Universal Gas constant
T = 273 k
`n = m/M`
m = mass of gas [g], given 10 g
M = mass of 1 mole of gas [g] that is 28.01 of CO

By putting all values -

`1*V=10/28.01*0.08206*273`
or
`V = 24.238/28.01`
or
V = 8.00

Hence Answer is 8.00 Liter Volume