Volume percentage of ethanol (C₂H₅OH) in its aqueous solution is 20%. If density of the solution is 0.96g/cc, calculate the molarity, molality of solution and mole fraction of each component ?
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Answers
Answer :-
→ Molarity of the solution = 3.47 M
→ Molality of the solution = 4.33 m
→ Mole fraction (ethanol) = 0.072
→ Mole fraction (water) = 0.928
Explanation :-
We have :-
• Concentration of ethanol = 20 %
• Density (solution) = 0.96 g/cc = 0.96 g/mL
• Density of water = 1 g/mL
• Molar mass of ethanol = 16 g/mol
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Let us assume that the volume of the solution is '100 mL' .
So, volume of ethanol (solute) is 20 mL and volume of water (solvent) is 80 mL .
Mass of the solution :-
= Volume × Density
= 100 × 0.96
= 96 g
Mass of water :-
= Volume × Density
= 80 × 1
= 80 g
Mass of Ethanol :-
= Mass of solution - Mass of water
= 96 - 80
= 16 g
Now, moles of ethanol present :-
= Given Mass/Molar mass
= 16/46
= 0.347 mole
Moles of water :-
= Given Mass/Molar mass
= 80/18
= 4.44 mole
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Molarity of the solution :-
= Moles of solute/Liters of solution
= 0.347/(100/1000)
= 0.347/0.1
= 3.47 M
Molality of the solution :-
= Moles of solute/Mass of solvent in kg
= 0.347/0.08
= 4.33 m
Mole fraction of ethanol :-
= Moles of ethanol/Total moles
= 0.347/(0.347 + 4.44)
= 0.347/4.787
= 0.072
Mole fraction of water :-
= Moles of water/Total moles
= 4.44/(0.347 + 4.44)
= 4.44/4.787
= 0.928
Given :-
Volume percentage of ethanol (C₂H₅OH) in its aqueous solution is 20%. If density of the solution is 0.96g/cc
To Find :-
Calculate the molarity, molality of solution and mole fraction of each component
Solution :-
Denisty of solution = 0.96 g/ml
Now
Density = Mass × Volume
Density = 0.960 x 100
Density = 960/1000 × 100
Density = 96/1
Density = 96 gm
Now,
The mass of the water = Volume of water × Density of water
mass of the water = 80 × 1
mass of the water = 80 gm
Now
Mass of solution = Mass of water + Mass of ethanol
96 = 80 + Mass of ethanol
96 - 80 = Mass of ethanol
16 gm = Mass of ethanol
Now
Molar mass of ethanol in the given chemical equation C₂H₅OH
12 × 2 + 1 × 5 + 16 × 1 + 1 × 1
24 + 5 + 16 + 1
29 + 17
46 gm
No. of moles of ethanol = Mass/Molar mass
No. of moles of ethanol = 16/46
No. of moles of ethanol = 0.3 mole
No. of moles of water = Mass/Molar mass
No. of moles of water = 80/18
No. of moles of water = 4.44 moles
Now
We know that
1 l = 1000 ml
100 ml = 100/1000 = 0.1 l
Molality = number of moles/volume in litres
Molality = 0.348/0.1
Molality = 348/1 × 10/1000
Molality = 3480/1000
Molality = 3.5 m
Molality = number of moles of solute/mass of solvent in kg
Molality = 0.348/0.08
Molality = 348/8 × 100/1000
Molality = 34800/8000
Molality = 4.3 m