How many H atoms are in 3.4 g of C12H22O11?1. 6.0 × 10232. 1.3 × 10233. 3.8 × 10224. 6.0 × 1021

To determine the number of H atoms in 3.4 g of C12H22O11, we need to use the molar mass of C12H22O11 and the Avogadro's number.

1. Calculate the molar mass of C12H22O11:

   • C: 12.01 g/mol x 12 = 144.12 g/mol
   • H: 1.01 g/mol x 22 = 22.22 g/mol
   • O: 16.00 g/mol x 11 = 176.00 g/mol Total molar mass = 144.12 g/mol + 22.22 g/mol + 176.00 g/mol = 342.34 g/mol

2. Calculate the number of moles of C12H22O11: Moles = Mass / Molar mass Moles = 3.4 g / 342.34 g/mol = 0.00992 mol

3. Determine the number of H atoms: Since there are 22 H atoms in one molecule of C12H22O11, we can use the following conversion: 1 mol C12H22O11 = 22 mol H atoms 0.00992 mol C12H22O11 x (22 mol H atoms / 1 mol C12H22O11) = 0.21824 mol H atoms

4. Convert moles to atoms: 1 mol H atoms = 6.022 x 10^23 atoms 0.21824 mol H atoms x (6.022 x 10^23 atoms / 1 mol H atoms) = 1.313 x 10^23 atoms

Therefore, there are approximately 1.3 × 10^23 H atoms in 3.4 g of C12H22O11.