Calculate the mass in grams of 3.78×10^21 molecules of N2O5

Calculate the mass in grams of 3.78×10^21 molecules of N2O5 (molar mass of N2O5 = 108.02 g/mol)

The Correct Answer and Explanation is:

To calculate the mass of 3.78×10213.78 \times 10^{21}3.78×1021 molecules of N2O5\text{N}_2\text{O}_5N2​O5​, we’ll follow these steps:

Step 1: Convert the number of molecules to moles

We know that 1 mole of any substance contains Avogadro’s number of molecules, which is approximately 6.022×10236.022 \times 10^{23}6.022×1023 molecules.

To find the number of moles in 3.78×10213.78 \times 10^{21}3.78×1021 molecules, use the formula:Moles=Number of moleculesAvogadro’s number\text{Moles} = \frac{\text{Number of molecules}}{\text{Avogadro’s number}}Moles=Avogadro’s numberNumber of molecules​

Substituting the values:Moles of N2O5=3.78×10216.022×1023=6.28×10−3 mol\text{Moles of } \text{N}_2\text{O}_5 = \frac{3.78 \times 10^{21}}{6.022 \times 10^{23}} = 6.28 \times 10^{-3} \text{ mol}Moles of N2​O5​=6.022×10233.78×1021​=6.28×10−3 mol

Step 2: Calculate the mass using the molar mass

We are given the molar mass of N2O5\text{N}_2\text{O}_5N2​O5​ as 108.02 g/mol. Now, to find the mass, we multiply the number of moles by the molar mass:Mass=Moles×Molar mass\text{Mass} = \text{Moles} \times \text{Molar mass}Mass=Moles×Molar mass

Substituting the values:Mass=6.28×10−3 mol×108.02 g/mol\text{Mass} = 6.28 \times 10^{-3} \text{ mol} \times 108.02 \text{ g/mol}Mass=6.28×10−3 mol×108.02 g/molMass=0.679 g\text{Mass} = 0.679 \text{ g}Mass=0.679 g

Final Answer:

The mass of 3.78×10213.78 \times 10^{21}3.78×1021 molecules of N2O5\text{N}_2\text{O}_5N2​O5​ is approximately 0.679 grams.

Explanation:

The key concept in this calculation is the relationship between the number of molecules, moles, and molar mass. First, we convert the number of molecules to moles using Avogadro’s number. Then, knowing that 1 mole of a substance corresponds to its molar mass in grams, we calculate the total mass of the given molecules. This process is essential when working with molecular quantities and converting them to macroscopic measurements like grams.

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