The number of atoms of hydrogen present in 1.255 g of aluminum hydroxide, Al(OH)₃ is 2.89×10²² atoms
How to determine the mole of 1.255 g of Al(OH)₃
- Mass of Al(OH)₃ = 1.225 g
- Molar mass of Al(OH)₃ = 27 + 3(16 + 1) = 78 g/mol
- Mole of Al(OH)₃ =?
Mole = mass / molar mass
Mole of Al(OH)₃ = 1.255 / 78
Mole of Al(OH)₃ = 0.016 mole
How to determine the mole of hydrogen
1 mole of Al(OH)₃ contains 3 moles of H.
Therefore,
0.016 mole of Al(OH)₃ will contain = 0.016 × 3 = 0.048 mole of H
How to determine the number of atoms in 0.048 mole of hydrogen
From Avogadro's hypothesis
1 mole of hydrogen = 6.02×10²³ atoms
Therefore,
0.048 mole of hydrogen = (0.048 mole × 6.02×10²³ atoms) / 1 mole
0.048 mole of hydrogen = 2.89×10²² atoms
Thus, we can conclude that 2.89×10²² atoms of hydrogen are present in 1.255 g of Al(OH)₃
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