Question

Calculate the mole fraction of ethylene glycol in a solution containing 20 percent of glycol by mass

Answer 1

*****Answer*****


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







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Answer 2

Answer:

Required mole fraction is 0.932.

Explanation:

20% of C₂H₆O₂ by mass is present.

In means solution,

Mass

$Moles \: of \: C₂H6O₂ \: is \: \frac{20}{62} = 0.322$

of Ethylene is 20 g and Mass of water is 80 g

Now,

$Molar \: mass \: of \: C₂H6O₂ = 12 \times 2 + 1 \times 6 + 2 \times 16 = 62g \: {mol}^{ - 1}$

$Moles \: of \: water = \frac{80}{18} = 4.444 \: moles$

Mole fraction of Ethylene glycol =

$\frac{0.322}{0.322 + 4.444} = 0.068$

Mole fraction of water is 1-0.068= 0.932.