60g of acetic acid is present in 540g of water. Calculate mole fraction of each component.
Answers
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Answer:-
→ Mole fraction of acetic acid = 0.032
→ Mole fraction of water = 0.968
Explanation:-
It is clear that acetic acid is solute and water is solvent.
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• Molar mass of Carbon (C) = 12 g/mol
• Molar mass of Hydrogen (H) = 1 g/mol
• Molar mass of Oxygen (O) = 16 g/mol
∴ Molar mass of acetic acid [CH₃COOH] :-
= 12 + 1×3 + 12 + 16 + 16 + 1
= 12 + 3 + 45
= 60 g/mol
∴ Molar mass of water [H₂O] :-
= 1×2 + 16
= 18 g/mol
Number of mole in 60g of acetic acid :-
= Given Mass/Molar mass
= 60/60
= 1 mole
Number of mole in 540g of water :-
= Given Mass/Molar mass
= 540/18
= 30 mole
Mole fraction of solute [acetic acid] :-
= Mole of solute/Total mole in solution
= 1/[30 + 1]
= 1/31
= 0.032
Mole fraction of solvent [water] :-
= Mole of solvent/Total mole in solution
= 30/[30 + 1]
= 30/31
= 0.968