Question

What is the concentration in molarity of an aqueous solution which contains 5.21% by mass ethylene glycol (mm=62.07g/mol)? The density of the solution is 1.04g/ml.

Answer 1

Answer:

C C2H6O2 = 0.873 M

Explanation:

• Molarity [=] mol / L

• wt% = (g C2H6O2 / g sln)×100

⇒ 0.0521 = g C2H6O2 / g sln

∴ mm C2H6O2 = 62.07 g/mol

∴ δ sln = 1.04 g/mL

assuming: g sln = 1 g

⇒ g C2H6O2 = 0.0521 g

⇒mol C2HJ6O2 = (0.0521 g)(mol/62.07 g) = 8.394 E-4 mol C2H6O2

⇒ V sln = 1g sln / 1.04 g/mL = 0.9615 mL = 9.615 E-4 L sln

⇒ C C2H6O2 = (8.394 E-4 mol)/(9.615 E-4 L sln)

⇒ C C2H6O2 = 0.873 mol/L

Answer 2

the concentration of the ethylene glycol in the solution is 0.87 M.

From the question given above, the following data were obtained:

Percentage by mass of ethylene glycol = 5.21%

Molar mass of ethylene glycol = 62.07 g/mol

Density = 1.04 g/mL

Concentration of ethylene glycol =?

Next, we shall mass ethylene glycol in the solution. This can be obtained as follow:

Percentage by mass of ethylene glycol = 5.21%

Therefore,

Mass of ethylene glycol = 5.21/100 = 0.0521 g

NOTE: the mass of the solution is assume to be 1 g

Next, we shall determine the number of mole in 0.0521 g of ethylene glycol.

Mass of ethylene glycol = 0.0521 g

Molar mass of ethylene glycol = 62.07 g/mol

Mole of ethylene glycol =?

Mole = mass / molar mass

Mole of ethylene glycol = 0.0521 / 62.07

Mole of ethylene glycol = 8.39×10¯⁴ mole

Next, we shall determine the volume of the solution.

Mass of solution = 1 g

Density of solution = 1.04 g/mL

Volume of solution =?

Density = mass /volume

1.04 = 1 / Volume

Cross multiply

1.04 × Volume = 1

Divide both side by 1.04

Volume = 1 / 1.04

Volume of solution = 0.962 mL

Divide by 1000 to express in litre ( L)

Volume of solution = 0.962 / 1000

Volume of solution = 9.62×10¯⁴ L

Finally, we shall determine the concentration of the ethylene glycol in the solution.

Mole of ethylene glycol = 8.39×10¯⁴ mole

Volume of solution = 9.62×10¯⁴ L

Concentration of ethylene glycol =?

Concentration = mole / Volume

Concentration of ethylene glycol = 8.39×10¯⁴ / 9.62×10¯⁴

Concentration of ethylene glycol = 0.87 M

Therefore, the concentration of the ethylene glycol in the solution is 0.87 M

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