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How many bromine atoms are present in 39.1 g of CH\(_2\)Br\(_2\)?

Answer :

Answer:

Answer:

2.77 × 10²³ Br Atoms


Solution:

Data Given:

Mass of CHâ‚‚Brâ‚‚ = 39.9 g


M.Mass of CHâ‚‚Brâ‚‚ = 173.83 g.mol⁻¹


Step 1: Calculate Moles of CHâ‚‚Brâ‚‚ as,


Moles = Mass ÷ M.Mass


Putting values,


Moles = 39.9 g ÷ 173.83 g.mol⁻¹


Moles = 0.23 mol


Step 2: Calculate number of CHâ‚‚Brâ‚‚ Molecules,


As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of CHâ‚‚Brâ‚‚ Molecules can be written as,


Moles = Number of CHâ‚‚Brâ‚‚ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹


Solving for Number of CHâ‚‚Brâ‚‚ Molecules,


Number of CHâ‚‚Brâ‚‚ Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹


Putting value of moles,


Number of CHâ‚‚Brâ‚‚ Molecules = 0.23 mol × 6.022 × 10²³ Atoms.mol⁻¹


Number of CHâ‚‚Brâ‚‚ Molecules = 1.38 × 10²³ CHâ‚‚Brâ‚‚ Molecules


Step 3: Calculate Number of Br Atoms:


As,


1 Molecule of CHâ‚‚Brâ‚‚ contains = 2 Atoms of Br


So,


1.38 × 10²³ Molecules of CHâ‚‚Brâ‚‚ will contain = X Atoms of Br


Solving for X,


X = (1.38 × 10²³ CHâ‚‚Brâ‚‚ × 2 Br) ÷ 1 CHâ‚‚Brâ‚‚


X = 2.77 × 10²³ Br Atoms

Explanation:

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Rewritten by : Jeany

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