# Mole fraction of ethanol in 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% THE ANSWER IS (C).... PLEASE GIVE A DETAILED SOLUTION

Aug 1, 2017

46%

#### Explanation:

I'll show you two methods that you can use to solve this problem.

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THE MORE TEDIOUS APPROACH
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As you know, a solution's percent concentration by mass tells you the number of grams of solute present for every $\text{100 g}$ of solution.

To make the calculations easier, pick a $\text{100-g}$ sample of this solution.

Now, you know that the mass of this sample will be equal to the mass of the ethanol, the solute, and the mass of the water, the solvent.

${m}_{\text{solution" = m_ "ethanol" + m_ "water}}$

In your case, you will have

${m}_{\text{ethanol" + m_ "water" = "100 g" " }} \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(1\right)}$

You also know that the mole fraction of ethanol, which is defined as the ratio between the number of moles of ethanol and the total number of moles present in the solution, is equal to $0.25$.

chi_ "ethanol" = n_"ethanol"/(n_"ethanol" + n_"water")

At this point, you must use the molar masses of ethanol and of water to express the mole ratio of ethanol in terms of ${m}_{\text{ethanol}}$ and ${m}_{\text{water}}$.

${M}_{\text{M ethanol" = "46.07 g mol}}^{- 1}$

${M}_{\text{M water" = "18.015 g mol}}^{- 1}$

This means that you have

${n}_{\text{ethanol" = m_"ethanol"/"46.07 g mol}}^{- 1}$

${n}_{\text{water" = m_"water"/"18.015 g mol}}^{- 1}$

Therefore, the mole fraction of ethanol can be rewritten as--for the sake of simplicity, I won't add any units

chi_ "ethanol" = (m_"ethanol"/46.07)/(m_"ethanol"/46.07 + m_"water"/18.015)

which is equivalent to

(18.015 * m_"ethanol")/(18.015 * m_"ethanol" + 46.07 * m_"water") = 0.25" "color(darkorange)((2))

Now all you have to do is to solve this system of two equations with two unknowns.

Use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(1\right)}$ to write

${m}_{\text{water" = 100 - m_"ethanol}}$

Plug this into equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(2\right)}$ to find

18.015 * m_"ethanol" = 0.25 * 18.015 * m_"ethanol" + 0.25 * 46.07 * (100 - m_"ethanol")

This will get you

${m}_{\text{ethanol}} \cdot \left(18.015 - 0.25 \cdot 18.015 + 0.25 \cdot 46.07\right) = 0.25 \cdot 46.07 \cdot 100$

which results in

${m}_{\text{ethanol}} = \frac{1151.75}{25.02875} = 46.02$

Since this represents the mass of ethanol present in $\text{100 g}$ of solution, you can say that the percent concentration by mass of ethanol is

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% ethanol by mass = 46%}}}}$

$\textcolor{w h i t e}{a}$
THE LESS TEDIOUS APPROACH
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Alternatively, you can start by picking a sample of this solution that contains exactly $1$ mole solute and of solvent.

This means that you have

${n}_{\text{ethanol" + n_"water" = "1 mole}}$

Now, you can use the mole fraction of ethanol to say that the number of moles of ethanol present in this sample is equal to

${\chi}_{\text{ethanol" = n_"ethanol"/"1 mole" implies n_"ethanol" = 0.25 * "1 mole" = "0.25 moles}}$

Consequently, you can say that this sample contains $0.75$ moles of water.

Use the molar masses of the two compounds to convert the number of moles to grams.

0.25 color(red)(cancel(color(black)("moles ethanol"))) * "46.07 g"/(1color(red)(cancel(color(black)("mole ethanol")))) = "11.52 g"

0.75 color(red)(cancel(color(black)("moles water"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole water")))) = "13.51 g"

The total mass of the solution will be

$\text{11.52 g + 13.51 g = 25.03 g}$

You can use the known composition of the sample to figure out how many grams of ethanol you'd ge for $\text{100 g}$ of this solution

100 color(red)(cancel(color(black)("g solution"))) * "11.52 g ethanol"/(25.03 color(red)(cancel(color(black)("g solution")))) = "46.02 g ethanol"

Once again, you have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% ethanol by mass = 46%}}}}$