Question

What is the mole fraction of each component in a solution made by mixing 300 g of ethanol(C2H5OH) and 500 g of water?

Answer 1

Answer:

The mole fraction of ethanol and water in the solution are 0.19 and 0.81 respectively.

Explanation:

The Molar Fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution, which are calculated by adding the moles of solute (s) and solvent.

It is calculated by dividing the number of moles of one of the components by the total number of moles of the solution.

The sum of all the molar fractions of the substances present in a solution is equal to 1.

So, the expression to calculate the mole fraction is:

$Molar fraction (Xi)=\frac{moles of sudstance}{total moles of the solution}$

Being the molar mass of the compounds:

• Ethanol 46 $\frac{g}{mol}$

• Water 18 $\frac{g}{mol}$

the number of moles that represent the mass quantities is calculated as:

• Moles of ethanol: 300 grams* $\frac{1 mol}{46 grams}$ = 6.52 moles
• Moles of water: 500 grams* $\frac{1 mol}{18 grams}$ = 27.78 moles

So the total moles in solution are 6.52 moles + 27.78 moles = 34.3 moles

The mole fraction of ethanol in the solution is calculated as:

$mole fraction of ethanol=x_{ethanol} =\frac{6.52 moles}{34.3 moles}$

$x_{ethanol} =0.19$

The mole fraction of water in the solution is calculated as:

$mole fraction of water=x_{water} =\frac{27.78 moles}{34.3 moles}$

$x_{water} =0.81$

Then:

$x_{water} +x_{ethanol} = 0.81 + 0.19$

$x_{water} +x_{ethanol} =1$

The mole fraction of ethanol and water in the solution are 0.19 and 0.81 respectively.

 

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