
1 mole of nitric acid produce 1 mole of ammonium nitrate.
moles in 5000 kg of ammonium nitrate :
( molecular mass of ammonium nitrate is 80 gm/mol )
So, number of moles of nitric acid required are also 62500 moles.
Mass of 62500 moles of nitric acid :

Hence, this is the required solution.
<h3>Answer:</h3>
7.57 × 10⁻²² g of F
<h3>Solution:</h3>
Data Given:
Number of Molecules = 8
M.Mass of BF₃ = 67.82 g.mol⁻¹
Mass of Fluorine atoms = ?
Step 1: Calculate Moles of BF₃
Moles = Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Putting value,
Moles = 8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Moles = 1.33 × 10⁻²³ mol
Step 2: Calculate Mass of BF₃:
Moles = Mass ÷ M.Mass
Solving for Mass,
Mass = Moles × M.Mass
Putting values,
Mass = 1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹
Mass = 9.0 × 10⁻²² g
Step 3: Calculate Mass of Fluorine Atoms:
As,
67.82 g BF₃ contains = 57 g of F
So,
9.0 × 10⁻²² g will contain = X g of F
Solving for X,
X = (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g
X = 7.57 × 10⁻²² g of F
B: quantity of solute with a specific volome of solvent.
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