Question:easy

How many molecules are present in $22400\ \mathrm{cm^3}$ of a gas at STP?

Show Hint

Keep unit equivalences in mind: $22.4\ \text{liters} = 22.4\ \mathrm{dm^3} = 22400\ \mathrm{cm^3} = 22400\ \mathrm{mL}$. Any gas with this volume at STP represents exactly one mole.
Updated On: Jun 11, 2026
  • $22.4 \times 10^{20}$
  • $6.022 \times 10^{23}$
  • $6.022 \times 10^{20}$
  • $22.4 \times 10^{23}$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Frame the problem.
We want the number of gas molecules in $22400\ \mathrm{cm^3}$ at STP. The key is to convert volume into moles, then moles into molecules.
Step 2: Recall the molar volume at STP.
One mole of any ideal gas occupies $22.4\ \mathrm{L}$ at STP, and $22.4\ \mathrm{L} = 22400\ \mathrm{cm^3}$ since $1\ \mathrm{L} = 1000\ \mathrm{cm^3}$.
Step 3: Find the number of moles.
\[ n = \frac{V}{V_m} = \frac{22400\ \mathrm{cm^3}}{22400\ \mathrm{cm^3\,mol^{-1}}} = 1\ \mathrm{mol}. \]
Step 4: Bring in Avogadro's constant.
Each mole holds $N_A = 6.022 \times 10^{23}$ particles.
Step 5: Multiply moles by $N_A$.
\[ N = n \times N_A = 1 \times 6.022 \times 10^{23} = 6.022 \times 10^{23}\ \text{molecules}. \]
Step 6: Conclude.
The sample contains $6.022 \times 10^{23}$ molecules, option (B). The volume was deliberately set to exactly one molar volume, so the count is just $N_A$ itself.
\[ \boxed{6.022 \times 10^{23}\ \text{molecules}} \]
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