Calculate the mole fraction of ethylene glycol $C_{2} H_{6} O_{2}$ in a solution containing $20 % of C_{2} H_{6} O_{2}$ by mass.

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Solution

Moles of compound

Given:

Mass percentage of ethylene glycol $(C_{2} H_{6} O_{2}) in a solution = 20 %$

i.e., $100 g$ of solution contains $20 g$ of ethylene glycol and $82 g$ of water.

Molar mass of $C_{2} H_{6} O_{2} = 12 \times 2 + 1 \times 6 + 16 \times 2 = 62 g m o l^{- 1}$

Moles of $C_{2} H_{6} O_{2} = \frac{mass of C_{2} H_{6} O_{2}}{Molar mass of C_{2} H_{6} O_{2}} = \frac{20 g}{62 g m o l^{- 1}}$
$= 0.322 m o l$

Similarly, moles of water $= \frac{80 g}{18 g m o l^{- 1}} = 4.444 m o l$

Mole fraction of ethylene glycol

$x_{glycol} = \frac{moles of C_{2} H_{6} O_{2}}{moles of C_{2} H_{6} O_{2} + moles of H_{2} O}$

$= \frac{0.322 m o l}{0.322 m o l + 4.444 m o l} = 0.068$

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