Question

If 110.4 grams of oxygen gas exerts 101.3 kPa of air pressure at 0°C, what volume would the gas occupy?

Answer 1

`154.69"L"`

Explanation:

We use the Ideal Gas Law:

`PV=nRT`

First we must convert the mass to moles. The molar mass of oxygen is `15.999"g/mol"`, so:

`n=110.4/15.999=6.9"mol"`

Here,

`P=101.3"kPa"`

`n=6.9"mol"`

`T=273.15"K"`

`R=8.314"L kPa K"^-1"mol"^-1`

Rearranging the equation to solve for `V`:

`V=(nRT)/P` and inputting:

`V=(6.9*8.314*273.15)/101.3`

`V=154.69"L"`