Which has the greater number of molecules: 10 g of \(N_{2}\) or 10 g of \(\mathrm{O}_{2} ?\)

Short Answer

Expert verified
10 g of \(N_{2}\) has a greater number of molecules than 10 g of \(\mathrm{O}_{2}\).

Step by step solution

01

Calculate the Moles of \(N_{2}\)

Firstly, calculate the moles of \(N_{2}\). The molar mass of \(N_{2} = 14.0 \times 2 = 28.0 \, \mathrm{g/mol}\). So, the number of moles of \(N_{2} = \frac{10 \, \mathrm{g}}{28.0 \, \mathrm{g/mol}} = 0.357 \, \mathrm{moles }\)_
02

Calculate the Moles of \(\mathrm{O}_{2}\)

Secondly, calculate the moles of \(\mathrm{O}_{2}\). The molar mass of \(\mathrm{O}_{2} = 16.0 \times 2 = 32.0 \, \mathrm{g/mol}\). So, the number of moles of \(\mathrm{O}_{2} = \frac{10 \, \mathrm{g}}{32.0 \, \mathrm{g/mol}} = 0.313 \, \mathrm{moles }\)_
03

Compare the Number of Moles

Finally, compare the number of moles. The number of molecules corresponds to the number of moles. Since \(N_{2}\) has 0.357 moles while \(\mathrm{O}_{2}\) has 0.313 moles, we can conclude that 10 g of \(N_{2}\) has more molecules than 10 g of \(\mathrm{O}_{2}\). The greater the number of moles, the greater the number of molecules, because 1 mole always contains a fixed number of entities (\(6.022 \times 10^{23}\), Avogadro's number).

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