Question

How many moles of argon atoms are present in 11.2 L of argon gas at STP?

Answer 1

What is the molar volume at `"STP"?`

Explanation:

Well, the molar volume at `"STP"` is `22.4*L`. You have to be careful to quote this now, because different curricula have different standards, and some quote standard pressure as `100*kPa` not `"1 atmosphere"`.

So you have a volume of `11.2*L`; given the molar volume, and this represents, `(11.2*cancelL)/(22.4*cancelL*mol^-1)=??mol`.

We note that `1/(mol^-1)=1/(1/(mol))=mol`, as required for a molar quantity.

Answer 2

Using the current recommended STP values of `"0"^@"C"` and `"100 kPa"`, `"11.2 L"` of Ar atoms contain `"0.493 mol Ar"`.

Using the older standard, as indicated in the explanation, `"11.2 L"` of Ar atoms contain `"0.500 mol"`.

Explanation:

Current values for STP are `0^@"C"` or `"273.15 K"` for temperature, and pressure of `"10"^5color(white)(.) "Pa"`, which is usually written as `"100 kPa"`, or `"1 bar"`.

At these current values, the molar volume of a gas is `"22.710 L/mol"`.

In order to determine the number of moles of the given volume of Ar, divide the given volume by the molar volume of an ideal gas, `"22.710 L/mol"`.

`(11.2"L")/(22.710"L"/"mol")`

Simplify by multiplying by the inverse of the molar volume.

`11.2cancel"L"xx(1"mol")/(22.710cancel"L")="0.493 mol Ar"`

Using the older STP values of `"0"^@"C"` or `"273.15 K"`, and pressure of `"1 atm"`, molar volume is `"22.414 L/mol"`.

`11.2cancel"L"xx(1"mol")/(22.414cancel"L")="0.500 mol Ar"`