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

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

Answer:

Given Values

Weth = 20 gm

Meth = C2H6O2 = 2(12)+6(1)+2(16) = 62

Wwater = 80 gm

Mwater = H2O = 2(1)+16 = 18

Calculations

χeth = \( \frac{n_{eth}}{n_{eth}+n_{water}} \)

Moles of Ethylene glycol

\( n_{eth} = \frac{W_{eth}}{M_{eth}} = \frac{20}{62} \) = 0.32

Moles of water

\( n_{water} = \frac{W_{water}}{M_{water}} = \frac{80}{18} \) = 4.44

Calculate mole fraction of ethylene glycol

χeth = \( \frac{0.32}{0.32+4.44} = 0.067 \)

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