# 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

A: 25%

B: 75%

C: 46%

D: 54%

THE ANSWER IS (C).... PLEASE GIVE A DETAILED SOLUTION

##### 1 Answer

#### Explanation:

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

**THE MORE TEDIOUS APPROACH**

As you know, a solution's **percent concentration by mass** tells you the number of grams of solute present *for every*

To make the calculations easier, pick a

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_ "solution" = m_ "ethanol" + m_ "water"#

In your case, you will have

#m_ "ethanol" + m_ "water" = "100 g" " "color(darkorange)((1))#

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

#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_ "M ethanol" = "46.07 g mol"^(-1)#

#M_ "M water" = "18.015 g mol"^(-1)#

This means that you have

#n_"ethanol" = m_"ethanol"/"46.07 g mol"^(-1)#

#n_"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

#m_"water" = 100 - m_"ethanol"#

Plug this into equation

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

This will get you

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

which results in

#m_"ethanol" = 1151.75/25.02875 = 46.02#

Since this represents the mass of ethanol present in

#color(darkgreen)(ul(color(black)("% ethanol by mass = 46%")))#

**THE LESS TEDIOUS APPROACH**

Alternatively, you can start by picking a sample of this solution that contains **exactly** **mole** solute *and* of solvent.

This means that you have

#n_"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_"ethanol" = n_"ethanol"/"1 mole" implies n_"ethanol" = 0.25 * "1 mole" = "0.25 moles"#

Consequently, you can say that this sample contains **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

#"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

#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

#color(darkgreen)(ul(color(black)("% ethanol by mass = 46%")))#