Question

Mole fraction of ethanol and water mixture is 0.25. Hence, percentage concentration of ethanol by weight of mixture is:
A. 25%
B. 75%
C. 46%
D. 54%

Answer 1

Hint: Mole fraction of ethanol (${C_2}{H_6}O$) and water (${H_2}O$), is determined by knowing the weight of 0.25 ethanol and total weight of the ethanol.

Step by step answer:
We know, the molecular formula of ethanol of ${C_2}{H_6}O$
Molar weight = Sum of the atomic masses of the constituent elements.
Atomic mass of C is $12g/mole$
Atomic mass of H is $1g/mole$
Atomic mass of O is $16g/mole$
Molar weight of Ethanol is $= {\text{ }}2 \times 12 + 6 \times 1 + 16{\text{ }} = 46{\text{ }}g/mole$
So 0.25 mole of Ethanol have weight of $0.25 \times 46{\text{ }} = {\text{ }}11.6{\text{ }}g$
Similarly,
Molar weight of ${H_2}O$ is $2 \times 1{\text{ }} + {\text{ }}16{\text{ }} = {\text{ }}18{\text{ }}g/mole$
So 0.75 mole of water have weight of $0.75 \times 18{\text{ }} = {\text{ }}13.5{\text{ }}g$
Therefore, the total weight of ethanol is $11.6{\text{ }}g{\text{ }} + {\text{ }}13.5{\text{ }}g{\text{ }} = {\text{ }}25.1{\text{ }}g$
Now, the percentage concentration of ethanol is
$\left( {\dfrac{{11.6}}{{25.1}}} \right) \times 100{\text{ }} = {\text{ }}46.21\%$
$\approx 46\%$

So we can see that Option C is the correct answer.

NOTE: Pure Ethanol is flammable, colourless liquid with a boiling point of $78.5^\circ C$. It has a low melting point of $- 144.5^\circ C$. This property of ethanol allows it to be used as antifreeze products.
If we have a mixture of N components and let us take moles of each component as
 ${{\text{n}}_{\text{1}}}{\text{,}}{{\text{n}}_{\text{2}}}{\text{,}}{{\text{n}}_{\text{3}}}{\text{, \ }}{{\text{n}}_{\text{N}}}$
Also, let us take the molar masses of each component as
${{\text{M}}_{\text{1}}}{\text{,}}{{\text{M}}_{\text{2}}}{\text{,}}{{\text{M}}_{\text{3}}}{\text{, \ }}{{\text{M}}_{\text{N}}}$
Therefore, the mole fraction of any component is:
${{\text{x}}_{\text{i}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{i}}}}}{{\sum\limits_{{\text{i = 1}}}^{\text{N}} {{{\text{n}}_{\text{i}}}} }}$
Therefore, the corresponding mass fraction can be written as:
${{\text{y}}_{\text{i}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{i}}}{{\text{M}}_{\text{i}}}}}{{\sum\limits_{{\text{i = 1}}}^{\text{N}} {{{\text{n}}_{\text{i}}}} {{\text{M}}_{\text{i}}}}}$
Hence, the corresponding mass percent can be simply written as $\left( {{\text{100 }} \times {\text{ }}{{\text{y}}_{\text{i}}}} \right)$