Question

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.

• A. 0.932
• B. 0.862
• C. 0.950
• D. 0.779

Answer 1

Answer: (A)

Assuming that we have $100g$ of solutions,
Mass of ethylene glycol $=20g$
and mass of water $=80g$
Molar mass of $C_2{H}_6{O}_2=12\times 2+1\times 6+16\times 2=62gmo{l}^{-1}$.
Moles of $C_2{H}_6{O}_2$ $=\frac{20g}{62gmo{l}^{-1}}=0.322mol$
Moles of water $=\frac{80g}{18gmo{l}^{-1}}=4.444mol$
$x_{glycol}=\frac{moleof{C}_2{H}_6{O}_2}{moleof{C}_2{H}_6{O}_2+moleof{H}_{2}O}$
$=\frac{0.322mol}{0.322mol+4.444mol}=0.068$
Similarly, $x_{water}=\frac{4.444mol}{0.322mol+4.444mol}=0.932$
Mole fraction of water can also be calculated as : $1-0.068=0.932$