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Question: A solution is \(25\% \) water, \(25\% \) ethanol and\(50\% \) acetic acid by mass. Calculate the mol...

A solution is 25%25\% water, 25%25\% ethanol and50%50\% acetic acid by mass. Calculate the mole fraction of each component.

Explanation

Solution

Mole fraction of the component is calculated by dividing the mole of the component to the total moles of all the components in the solution. The moles of a component is calculated by dividing the given mass of the component to the molecular mass of it. We shall find the number of moles present in the solution from the percentage by assuming the total mass of solution as 100g. Then, the mole fraction will be calculated using the formula (given below).
Formula used
moles = given massmolecular mass{\text{moles = }}\dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}
Mole fraction of solute = x2 = n2n2 + n1{\text{Mole fraction of solute = }}{{\text{x}}_{\text{2}}}{\text{ = }}\dfrac{{{{\text{n}}_{\text{2}}}}}{{{{\text{n}}_{\text{2}}}{\text{ + }}{{\text{n}}_{\text{1}}}}}
Mole fraction of solvent = x1 = n1n2 + n1{\text{Mole fraction of solvent = }}{{\text{x}}_{\text{1}}}{\text{ = }}\dfrac{{{{\text{n}}_1}}}{{{{\text{n}}_{\text{2}}}{\text{ + }}{{\text{n}}_{\text{1}}}}}
Where x1 and x2{{\text{x}}_{\text{1}}}{\text{ and }}{{\text{x}}_{\text{2}}} are mole fractions of solvent and solute and n1 and n2{{\text{n}}_1}{\text{ and }}{{\text{n}}_2} are number of moles of solvent and solute

Complete step by step answer:
So, in the question the solution contains 25%25\% water, 50%50\% ethanoic acid, and 25%25\% ethanol by mass. 25%25\% of water means 25g of water is present in 100g of solution. 50% of ethanoic acid means 50g of ethanoic acid is present in 100g of solution, 25%25\% of ethanol means 25g of ethanol is present in 100g of solution.
Molar mass of water is 18 gmol1{\text{gmo}}{{\text{l}}^{ - 1}} , molar mass of ethanol is 46 gmol1{\text{gmo}}{{\text{l}}^{ - 1}} , molar mass of ethanoic acid is 60 gmol1{\text{gmo}}{{\text{l}}^{ - 1}}
Number of moles of water=2518=1.4{\text{Number of moles of water}} = \dfrac{{25}}{{18}} = 1.4
  Number of moles of ethanoic acid=5060=0.8\;{\text{Number of moles of ethanoic acid}} = \dfrac{{50}}{{60}} = 0.8
Number of moles of ethanol=2546=0.54{\text{Number of moles of ethanol}} = \dfrac{{25}}{{46}} = 0.54
So, the total number of moles of all components in the solution =1.4+0.54+0.8=2.74 = 1.4 + 0.54 + 0.8 = 2.74
So, we have the moles of each component and total moles of the solution, therefore,
  Mole fraction of water=1.42.74=0.51\;{\text{Mole fraction of water}} = \dfrac{{1.4}}{{2.74}} = 0.51
Mole fraction of ethanoic acid=0.82.74=0.291{\text{Mole fraction of ethanoic acid}} = \dfrac{{0.8}}{{2.74}} = 0.291
Mole fraction of ethanol=0.542.74=0.197{\text{Mole fraction of ethanol}} = \dfrac{{0.54}}{{2.74}} = 0.197

Hence the mole fraction of water is0.510.51, ethanoic acid is 0.2910.291 , and ethanol is 0.1970.197.

Note:

Mole fraction is a dimensionless quantity because we are dividing the moles with moles. When the percentage of the components is converted into mass, 100g100{\text{g}} of the solution is taken for the easy calculation.