Calculate the mole fraction of ethylene glycol -C2H6O2- in a solution containing 20

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Standard XII

Chemistry

Abnormal Colligative Properties

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Question

Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂by mass.

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Solution

Mass of ethylene glycol in 100g of solution= 20g
Mass of water in 1oo g solution= 80g.
Number of moles of ethylene glycole= 20/69 =0.2
Number of moles of water =80/18 =4.44
Therefore,
Mole fraction of ethylene glycol = 0.2/(4.44+0.2) = 0.04

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Q. An antifreeze mixture consists of

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QUESTION 2.8

An antifreeze solution is prepared from 222.6 g of ethylene glycol $(C_{2} H_{6} O_{2})$ and 200 g of water. Calculate the molality of the solution. If the density of the solution is $1.072 g m L^{- 1}$ , then what shall be the molarity of the solution?

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Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing 20% of C2H6O2 by mass.

Mass of ethylene glycol in 100g of solution= 20gMass of water in 1oo g solution= 80g.Number of moles of ethylene glycole= 20/69 =0.2Number of moles of water =80/18 =4.44Therefore,Mole fraction of ethylene glycol = 0.2/(4.44+0.2) = 0.04

Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂by mass.

Mass of ethylene glycol in 100g of solution= 20g
Mass of water in 1oo g solution= 80g.
Number of moles of ethylene glycole= 20/69 =0.2
Number of moles of water =80/18 =4.44
Therefore,
Mole fraction of ethylene glycol = 0.2/(4.44+0.2) = 0.04