Bettu
Jun 19
What is the volume of acetic acid that is 11% impure acid and has density of 2.4g/cm3 is needed to prepare 1 litre of 0.5M of Acetic acid solution?( The molar mass of acetic acid is 60.06g/mol.)?
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ChatGPT
Jun 19
To calculate the volume of pure acetic acid needed to prepare 1 litre of 0.5M acetic acid solution, we first need to determine the amount of pure acetic acid in the 11% impure acetic acid.
Let x be the volume of the impure acetic acid needed.
Given:
Density of impure acetic acid = 2.4 g/cm^3
Percentage of pure acetic acid in impure acid = 11%
Molar mass of acetic acid = 60.06 g/mol
First, we need to calculate the mass of pure acetic acid in x volume of impure acetic acid:
Mass of pure acetic acid = Volume x Density
Mass of pure acetic acid = x cm^3 x 2.4 g/cm^3
Mass of pure acetic acid = 2.4x g
Since the impure acetic acid is 11% pure, the mass of pure acetic acid in x volume of impure acetic acid is:
0.11(2.4x) = 0.264x g
Now, we need to calculate the number of moles of pure acetic acid in x volume of impure acetic acid:
Number of moles = Mass / Molar mass
Number of moles = 0.264x / 60.06
Now, we know that the molarity (M) is given by:
M = Number of moles / Volume (in litres)
0.5 = 0.264x / (60.06 x 1000) (since 1 litre = 1000 cm^3)
Solving for x:
x = (0.5 x 60.06 x 1000) / 0.264
x = 11363.64 cm^3
Therefore, the volume of the impure acetic acid needed is 11363.64 cm^3.
To prepare 1 litre of 0.5M acetic acid solution, we need to dilute this impure acetic acid with water.
Let x be the volume of the impure acetic acid needed.
Given:
Density of impure acetic acid = 2.4 g/cm^3
Percentage of pure acetic acid in impure acid = 11%
Molar mass of acetic acid = 60.06 g/mol
First, we need to calculate the mass of pure acetic acid in x volume of impure acetic acid:
Mass of pure acetic acid = Volume x Density
Mass of pure acetic acid = x cm^3 x 2.4 g/cm^3
Mass of pure acetic acid = 2.4x g
Since the impure acetic acid is 11% pure, the mass of pure acetic acid in x volume of impure acetic acid is:
0.11(2.4x) = 0.264x g
Now, we need to calculate the number of moles of pure acetic acid in x volume of impure acetic acid:
Number of moles = Mass / Molar mass
Number of moles = 0.264x / 60.06
Now, we know that the molarity (M) is given by:
M = Number of moles / Volume (in litres)
0.5 = 0.264x / (60.06 x 1000) (since 1 litre = 1000 cm^3)
Solving for x:
x = (0.5 x 60.06 x 1000) / 0.264
x = 11363.64 cm^3
Therefore, the volume of the impure acetic acid needed is 11363.64 cm^3.
To prepare 1 litre of 0.5M acetic acid solution, we need to dilute this impure acetic acid with water.
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Google Gemini AI
Jun 19
Step 1: Calculate the mass of pure Acetic acid needed
Molarity (M) = Moles of solute / Volume of solution (in liters)
0.5 M = Moles of acetic acid / 1 L
Moles of acetic acid = 0.5 mol
Mass of pure acetic acid = Moles of acetic acid x Molar mass
= 0.5 mol x 60.06 g/mol
= 30.03 g
Step 2: Calculate the mass of 11% impure acetic acid needed
Mass of pure acetic acid / Mass of impure acid = Purity
30.03 g / Mass of impure acid = 0.11
Mass of impure acid = 30.03 g / 0.11
= 273.0 g
Step 3: Calculate the volume of impure acetic acid needed
Density = Mass / Volume
2.4 g/cm³ = 273.0 g / Volume
Volume = 273.0 g / 2.4 g/cm³
= 113.75 cm³
Therefore, 113.75 cm³ of 11% impure acetic acid with a density of 2.4 g/cm³ is needed to prepare 1 liter of 0.5 M acetic acid solution.
Molarity (M) = Moles of solute / Volume of solution (in liters)
0.5 M = Moles of acetic acid / 1 L
Moles of acetic acid = 0.5 mol
Mass of pure acetic acid = Moles of acetic acid x Molar mass
= 0.5 mol x 60.06 g/mol
= 30.03 g
Step 2: Calculate the mass of 11% impure acetic acid needed
Mass of pure acetic acid / Mass of impure acid = Purity
30.03 g / Mass of impure acid = 0.11
Mass of impure acid = 30.03 g / 0.11
= 273.0 g
Step 3: Calculate the volume of impure acetic acid needed
Density = Mass / Volume
2.4 g/cm³ = 273.0 g / Volume
Volume = 273.0 g / 2.4 g/cm³
= 113.75 cm³
Therefore, 113.75 cm³ of 11% impure acetic acid with a density of 2.4 g/cm³ is needed to prepare 1 liter of 0.5 M acetic acid solution.
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