Question

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calculate the molarity of 20% of aqueous solution of ethylene glycol [ C2H6O2 ].

please answer​

Answer 1

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Explanation:20% of C2H6O2 by mass is present.

That means solution has 20 g of ethylene glycol and 80 g of water.

Now,

Molar mass of C2H6O2 = 12 x 2 + 1 x 6 + 2 x 16 = 62 g mol-1 .

Moles of C2H6O2 = 20 / 62 = 0.322 moles

Moles of water = 80 / 18 = 4.444 moles

Mole fraction of ethylene glycol = 0.322 / 0.322 + 4.444 = 0.068

Mole fraction of water = 1 - 0.068 = 0.932

Answer 2

Answer:

Explanation:Explanation:20% of C2H6O2 by mass is present.

That means solution has 20 g of ethylene glycol and 80 g of water.

Now,

Molar mass of C2H6O2 = 12 x 2 + 1 x 6 + 2 x 16 = 62 g mol-1 .

Moles of C2H6O2 = 20 / 62 = 0.322 moles

Moles of water = 80 / 18 = 4.444 moles

Mole fraction of ethylene glycol = 0.322 / 0.322 + 4.444 = 0.068

Mole fraction of water = 1 - 0.068 = 0.932