\(5.49 \times 10^{24}\) particles of ethyne are present in 238 grams of ethyne.
To determine the number of particles in 238 grams of ethyne (C2H2), we need to use Avogadro's number and the molar mass of ethyne.
1. Find the moles of ethyne using the given mass:
\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{moles} = \frac{238 \, \text{g}}{26.038 \, \text{g/mol}} \]
\[ \text{moles} \approx 9.14 \, \text{mol} \]
2. Apply Avogadro's number (6.022 x 10^23 particles/mol) to convert moles to particles:
\[ \text{particles} = \text{moles} \times \text{Avogadro's number} \]
\[ \text{particles} = 9.14 \, \text{mol} \times 6.022 \times 10^{23} \, \text{particles/mol} \]
\[ \text{particles} \approx 5.49 \times 10^{24} \]
Final Answer: \(5.49 \times 10^{24}\) particles of ethyne are present in 238 grams of ethyne.