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How many oxygen atoms are there in 4.00 g of sodium dichromate, Na₂Cr₂O₇?

A. 6.44 x 10²² oxygen atoms
B. 9.19 10²¹ oxygen atoms
C. 0.107 oxygen atoms
D. 1.31 10²¹ oxygen atoms

User Quarkonia
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1 Answer

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Final answer:

To find the number of oxygen atoms in 4.00 g of sodium dichromate, we calculate the moles of Na₂Cr₂O₇ and then use Avogadro's number to convert it to atoms, resulting in approximately 6.44 × 10²² oxygen atoms.

Step-by-step explanation:

To determine the number of oxygen atoms in 4.00 g of sodium dichromate, Na₂Cr₂O₇, we must first calculate the number of moles of sodium dichromate in the sample using its molar mass. The molar mass of Na₂Cr₂O₇ is: (2 × 22.99) + (2 × 52.00) + (7 × 16.00) = 261.97 g/mol. Therefore, the number of moles of Na₂Cr₂O₇ in 4.00 g is calculated as follows:

Number of moles = mass (g) / molar mass (g/mol) = 4.00 g / 261.97 g/mol ≈ 0.0153 mol.

Each mole of Na₂Cr₂O₇ contains 7 moles of oxygen atoms. So, the total number of moles of oxygen atoms in 4.00 g of Na₂Cr₂O₇ is:

0.0153 mol × 7 = 0.1071 mol of oxygen atoms.

To find the total number of oxygen atoms, we multiply the number of moles of oxygen by Avogadro's number (6.022 × 10²³ atoms/mol):

0.1071 mol × 6.022 × 10²³ atoms/mol ≈ 6.44 × 10²² oxygen atoms.

Therefore, the correct answer is A. 6.44 x 10²² oxygen atoms.

User Jserras
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