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.

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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It decreases by about 10%

Explanation: