GIVEN THAT 32.0g SULPHUR CONTAINS 6.02 X 10^23 SULPHUR ATOMS, HOW MANY ATOMS ARE THERE IN 2.70g OF ALUMINIUM

• A. 6.02 x 10 ²³
• B. 3.01 x 10²²
• C. 6.02 x 10^{22 ✓}
• D. 5. 08 x 10²²
• E. 3.01 x 10²²

 

The answer to the question is: C. 6.02 x 10²²

To calculate the number of atoms in a sample, you can use Avogadro’s number and the molar mass of the element in question. Avogadro’s number, 6.02 x 10^23, is the number of atoms in one mole of any substance. The molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol).

Firstly, 32.0 g of sulfur contains 6.02 x 10^23 sulfur atoms. The molar mass of sulfur (S) is approximately 32.065 g/mol (this value can be found on the periodic table). Since the given mass of sulfur (32.0 g) almost equals its molar mass (32.065 g/mol), it means we have nearly one mole of sulfur, which contains Avogadro’s number of atoms.

Now let’s calculate for aluminum (Al). The molar mass of aluminum is approximately 26.98 g/mol.

To find out how many moles are in 2.70 g of aluminum, we use the formula:

Number of moles = Mass (g) / Molar mass (g/mol)

For aluminum:

Number of moles = 2.70 g / 26.98 g/mol ≈ 0.1 mol

Since one mole contains Avogadro’s number of atoms, to find the total number of atoms in 2.70 g of aluminum, we multiply the number of moles by Avogadro’s number:

Number of atoms = Number of moles × Avogadro’s number

Number of atoms in 2.70 g Al ≈ 0.1 mol × 6.02 x 10^23 atoms/mol ≈ 6.02 x 10^22 atoms

So, there are approximately 6.02 x 10^22 atoms in a 2.70 g sample of aluminum.