Question

$21\text{g}$ of nitrogen gas is mixed with 1.2 moles of $S{O}_2$ gas and $1.21\times {10}^{23}$ molecules of $O_2$ gas. Find the volume of the gaseous mixture (assuming no reaction) at STP.

• A. $22.4\text{L}$
• B. $15.67\text{L}$
• C. $48.16\text{L}$
• D. $67.56\text{L}$

Answer 1

The correct option is C $48.16\text{L}$

Moles of nitrogen $=\frac{21}{28}=0.75$ mole
1 mole of $O_2$ gas contains $6.023\times {10}^{23}$ molecules
So, $1.21\times {10}^{23}$ are present in 0.2 moles.
Total moles of gases $=0.75+1.2+0.2=2.15$ moles.
1 mole occupies $22.4\text{L}$ of volume
2.15 mole occupies $=22.4\times 2.15=48.16\text{L}$