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Chapter 3: Problem 9

Without doing detailed calculations, explain which of the following has the greatest number of \(\mathrm{N}\) atoms (a) \(50.0 \mathrm{g}\) \(\mathrm{N}_{2} \mathrm{O} ;\) (b) \(17.0 \mathrm{g} \mathrm{NH}_{3} ;\) (c) \(150 \mathrm{mL}\) of liquid pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(d=0.983 \mathrm{g} / \mathrm{mL}) ;\) (d) \(1.0 \mathrm{mol} \mathrm{N}_{2}\)

Short Answer

Expert verified
The substance with the greatest number of Nitrogen atoms is option (d) that is \(1.0 \, \mathrm{mol}\) of \(N_{2}\), because it contains 2 moles of Nitrogen atoms.

Step by step solution

01

Determine the Molecular Mass of each option

In this first step, the molecular masses are determined. For (a) N2O, the molecular mass is \(2(14.01) + 16.00 = 44.02 \, \mathrm{g/mol}\). For (b) NH3, the molecular mass is \(14.01 + 3(1.01) = 17.04 \, \mathrm{g/mol}\). For (c) C5H5N, the molecular mass is \(5(12.01) + 5(1.01) + 14.01 = 79.10 \, \mathrm{g/mol}\). Finally, for (d) N2, the molecular mass is \(2(14.01) = 28.02 \, \mathrm{g/mol}\).
02

Calculating the number of atoms for each option

In this step, we estimate which option gives the most number of Nitrogen atoms. For (a) \(50.0 \, \mathrm{g}\) of \(N2O\), it's approximately 1 mole of Nitrogen atoms. For (b) \(17.0 \, \mathrm{g}\) of \(NH3\), it's approximately 1 mole of Nitrogen atoms. For (c) \(150 \, \mathrm{mL}\) of pyridine \(C5H5N\), using the given density we can find its 'nitrogen mass' and using that it's again around 1 mole of Nitrogen atoms. For (d) \(1.0 \, \mathrm{mol}\) of \(N2\), we have 2 moles of Nitrogen atoms.
03

Drawing Conclusion

On comparing the moles of Nitrogen atoms from each option, it's clear that option (d) gives the highest number of Nitrogen atoms because it has 2 moles of Nitrogen atoms and all other choices have approximately 1 mole of Nitrogen atoms.

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