How many molecules of ammonia gas are present in 67.2 $\text{dm}^3$, measured at S.T.P.?
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Recognize common multiples of $22.4$ to speed up calculations! Notice that $22.4 \times 3 = 67.2$. Finding out that you have exactly 3 moles instantly simplifies the calculation to $3 \times 6 \times 10^{23} = 18 \times 10^{23} = 1.8 \times 10^{24}$.
Step 1: Spot the gas-at-STP clue. Any ideal gas occupies a fixed molar volume of $22.4$ dm$^3$ at STP. So volume can be turned straight into moles. Step 2: Recall Avogadro's number. One mole contains $N_A = 6.022 \times 10^{23}$ particles, so moles can be turned into a molecule count. Step 3: Write the plan as one chain. $\text{molecules} = \dfrac{\text{volume}}{22.4} \times N_A$. Step 4: Convert the given volume to moles. $n = \dfrac{67.2}{22.4} = 3$ mol. The numbers divide cleanly because $22.4 \times 3 = 67.2$. Step 5: Convert moles to molecules. Number of molecules $= 3 \times 6.022 \times 10^{23} = 18.066 \times 10^{23}$. Step 6: Express in proper form. $18.066 \times 10^{23} = 1.8 \times 10^{24}$ molecules, matching option (3). \[ \boxed{1.8 \times 10^{24}\ \text{molecules}} \]