Question

The active mass of 5.6 litres of O2 at STP is

Answer 1

Moles = 5.6/22.4 = 0.25 moles  Moles = mass / molecular mass 0.25 = mass / 32 Mass = 0.25*32 Mass = 8 grams

Answer 2

Answer: The active mass of oxygen gas at STP is 8 grams.

Explanation:

At STP conditions:

22.4 L of volume is occupied by 1 mole of a gas.

So, 5.6 L of volume will be occupied by = $\frac{1mol}{22.4L}\times 5.6L=0.25mol$

Now, to calculate the mass of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of oxygen gas = 0.25 mol

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.25mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=8g$

Hence, the active mass of oxygen gas at STP is 8 grams.