Question

22.4 L of hydrogen gas contains how many molecules?

Answer 1

At `0°"C"` temperature and `1` atmosphere pressure, this volume is one mole. This is Avogrado's number of molecules, `color(blue)(6.02xx10^{23})`.

Explanation:

The temperature and pressure chosen for this answer, `0°"C"` and `1` atmosphere, were intended to be "standard" temperature and pressure, and the question was apparently formulated with that standard in mind. But the standard pressure was changed to `100` kPa (one atmosphete is `101.3` kPa) in 1982, and confusingly some references still quote the old standard (e.g., http://www.kentchemistry.com/links/GasLaws/STP.htm). One solution to this dilemma that fits with the question is to avoid the "standard" label and refer to the temperature and pressure specifically.

The connection between `22.4" L"` and one mole, at `0°"C"` and `1` atmosphere, comes from the Ideal Gas Law

`PV=nRT`

`P` = pressure, `1` atm
`V` = volume, see below

`n` = moles of gas, put in `1` mole

`R` = gas constant, in pressure-volume unots it's `color(blue)(0.08206 ("L atm")/("mol K"))`

`T` = tempeature, `0°"C" = 273.15 "K"`

So

`(1" atm")(V)=(1" mol")(.08206 ("L atm")/("mol K"))(273.15" K")`

Solve for `V` and to three significant figures you get `color(blue)(22.4" L")`. Thus at `0°"C"` temperature and `1` atmosphere pressure, `22.4" L"` of an ideal gas is one mole.

Answer 2

At a pressure of 1 bar and a temperature of 0 °C, 22.4 L of hydrogen gas contains `5.94 × 10^23color(white)(l) "molecules"`.

Explanation:

The number of molecules depends on the temperature and pressure of the gas.

I shall arbitrarily assume that the gas is at STP (1 bar and 0 °C).

We can use the Ideal Gas Law to calculate the moles of hydrogen:

`color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))" "`

We can rearrange this to get

`n = (PV)/(RT)`

`P = "1 bar"`
`V = "22.4 L"`
`R = "0.0.08 314 bar·L·K"^"-1""mol"^"-1"`
`T = "(0 + 273.15) K" = "273.15 K"`

∴ `n = (1 color(red)(cancel(color(black)("bar"))) × 22.4 color(red)(cancel(color(black)("L"))))/("0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K")))) = "0.9864 mol"`

We know that 1 mol of a gas contains `6.022 × 10^23 color(white)(l)"molecules"`.

∴ `"No. of molecules" = 0.9864 color(red)(cancel(color(black)("mol"))) × (6.022 × 10^23color(white)(l) "molecules")/(1 color(red)(cancel(color(black)("mol")))) = 5.94 × 10^23color(white)(l) "molecules"`