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How many moles are in 1.25 x 1021 molecules of sucrose?

How many moles of aspartame are present in 197 g of aspartame, C14H18N2O5?

1 Answer

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Answer:

Therefore, there are approximately 0.669 moles of aspartame in 197 g of aspartame.

Step-by-step explanation:

To answer your first question, we need to use Avogadro's number, which is the number of particles in one mole of a substance. Avogadro's number is approximately 6.022 x 10^23 particles per mole. So, to find the number of moles in 1.25 x 10^21 molecules of sucrose, we divide the number of molecules by Avogadro's number:

1.25 x 10^21 molecules / 6.022 x 10^23 molecules per mole = 0.00208 moles of sucrose

Therefore, there are approximately 0.00208 moles of sucrose in 1.25 x 10^21 molecules of sucrose.

For your second question, we need to calculate the molar mass of aspartame and use it to convert the given mass to moles. The molar mass of aspartame, C14H18N2O5, can be calculated by summing the atomic masses of all the atoms in one molecule of aspartame:

Molar mass of aspartame = (14 x 12.011) + (18 x 1.008) + (2 x 14.007) + (5 x 15.999) = 294.3 g/mol

So, to find the number of moles of aspartame in 197 g, we divide the mass by the molar mass:

197 g / 294.3 g/mol = 0.669 moles of aspartame

Therefore, there are approximately 0.669 moles of aspartame in 197 g of aspartame.

User Thomas David Kehoe
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