Question

Calculate the mole fraction of ethylene glycol $(C_2H_6O_2)$ in a solution containing 20% of $C_2H_6O_2$ by mass.

• A. 0.932
• B. 0.862
• C. 0.95
• D. 0.779

Answer 1

Answer: (A)

0.932

Answer 2

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 : $1 - 0.068 = 0.932$