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Q.
Calculate the mole fraction of ethylene glycol $(C_2H_6O_2)$ in a solution containing 20% of $C_2H_6O_2$ by mass.
Solutions
Solution:
Assuming that we have $100\, g$ of solutions,
Mass of ethylene glycol $= 20 \,g$
and mass of water $= 80\, g$
Molar mass of $C_2H_6O_2 = 12 × 2 + 1 × 6 + 16 × 2 = 62 \,g\, mol^{-1}$.
Moles of $C_2H_6O_2$ $= \frac{20\,g}{62\,g\,mol^{-1}} = 0.322\,mol$
Moles of water $= \frac{80\,g}{18\,g\,mol^{-1}} = 4.444\,mol$
$x_{glycol} = \frac{mole\, of\, C_{2}H_{6}O_{2}}{mole\, of \,C_{2}H_{6}O_{2} + mole \,of\, H_{2}O}$
$= \frac{0.322\,mol}{0.322\,mol+4.444\,mol} = 0.068$
Similarly, $x_{water}= \frac{4.444\,mol}{0.322\,mol+4.444\,mol} = 0.932$
Mole fraction of water can also be calculated as : $1 - 0.068 = 0.932$