Question

A solution contains 25% water 25% ethanol and 50% acetic acid by mass calculate the mole fraction of each component

Answer 1

Let the mass of solution is 100g
then, mass of water = 25g
mass of ethanol = 25g
mass of acetic acid = 50g

Now, find number of moles of each components ,
Number of mole of water = mass of water/Molar mass of water
= 25g/18g/mol [ ∵ molar mass of water = 18g/mol ]
≈ 1.4

Number of moles of ethanol = mass of ethanol/molar mass of ethanol
= 25g/46g/mol [ ∵ molar mass of ethanol (C2H5OH) = 46g/mol
= 0.54

Number of moles of acetic acid = mass of acetic acid/molar mass of acetic acid
= 50g/60g/mol [∵ molar mass of acetic acid (CH3COOH) = 60g/mol]
= 0.8
∴ total number of moles = 1.4 + 0.54 + 0.8 = 2.74
Now, mole fraction of water = moles of water/total moles
= 1.4/2.74 =0.51

similarly , mole fraction of ethanol = 0.54/2.74 = 0.197
mole fraction of acetic acid = 0.8/2.74 = 0.291

Answer 2

Let us consider the total mass of the solution to be 100 g.

Thus, the mass of water will be 25% of 100 g, i.e. 25 g
Similarly, the mass of ethanol will be 25 g and
the mass of acetic acid will be 50 g.

Let us first calculate the number of moles of each component.

Molarity = Gram Mass / Molar Mass

We know that the molar mass of Water ($H_{2} O$) is 18, that of Ethanol ($C_{2} H_{5} OH$) is 46 and that of Acetic Acid ($CH_{3} COOH$) is 60.

Molarity of Water = 25/18 = 1.388 M
Molarity of Ethanol = 25/46 = 0.543 M
Molarity of Acetic Acid = 50/60 = 0.833 M

We know that Mole Fraction of a particular component = 
No. of moles of that component / Total no. of moles of all components in the solution

Total no. of moles = 1.388 + 0.543 + 0.833 = 2.764

Therefore, Mole fraction of Water = 1.388/2.764 = 0.502
Mole fraction of Ethanol = 0.543/2.764 = 0.196
Mole fraction of Acetic Acid = 0.833/2.764 = 0.301