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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.)?

Jun 19

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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.

Google Gemini AI

Jun 19

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

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

Density = Mass / Volume

2.4 g/cm³ = 273.0 g / Volume

Volume = 273.0 g / 2.4 g/cm³

= 113.75 cm³