Question

If 22g of ${\mathrm{CO}}_2$ occupy V litre at STP then what volume would 32g of ${\mathrm{O}}_2$ occupy at STP? Justify your answer.

Answer 1

Step 1: Molecular weight of ${\mathrm{CO}}_2$

The molecular weight of ${\mathrm{CO}}_2$ is

$$\begin{gathered} =12+(16\times 2) \\ =12+32 \\ =44\mathrm{g}/\mathrm{mol} \end{gathered}$$

Step 2: Number of mole for 22g of ${\mathrm{CO}}_2$

If 44g of ${\mathrm{CO}}_2$ corresponds to 1 mole, then 22g of ${\mathrm{CO}}_2$ will correspond to $\frac{22}{44}=\frac{1}{2}=0.5\mathrm{mole}$

Step 3: Molecular weight of ${\mathrm{O}}_2$

The molecular weight of oxygen is $(16\times 2)=32\mathrm{amu}$

Step 4:- Number of mole for 32g of ${\mathrm{O}}_2$:-

If 32g of ${\mathrm{O}}_2$ corresponds to 1 mole, then 32 of ${\mathrm{O}}_2$will correspond to =$\frac{32}{32}=1\mathrm{mole}$

Step 5: 1 mole of a substance at STP:-

We know that at STP 1mole of a substance will occupy 22.4L of volume, and here 32g of ${\mathrm{O}}_2$ corresponds 1mole.

Hence, 32g of ${\mathrm{O}}_2$ will occupy 22.4L of volume at STP.