Mole fraction of ethanol in ethanol and water mixture is 0.25. Hence percentage concentration of ethanol by weight of mixture is?

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

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.

In your case, you will have

This means that you have

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

which is equivalent to

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

This will get you

which results in

This means that you have

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

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

The total mass of the solution will be

Once again, you have

To find the percentage concentration of ethanol by weight in the mixture, you can use the formula:

[ \text{Percentage concentration by weight} = \left( \frac{\text{Mole fraction of component} \times \text{Molecular weight of component}}{\text{Mole fraction of component} \times \text{Molecular weight of component} + \text{Mole fraction of other component} \times \text{Molecular weight of other component}} \right) \times 100 ]

In this case, let ( x ) be the mole fraction of ethanol, and ( M_{\text{ethanol}} ) and ( M_{\text{water}} ) be the molecular weights of ethanol and water, respectively.

[ x = 0.25 ]

[ \text{Percentage concentration of ethanol by weight} = \left( \frac{0.25 \times M_{\text{ethanol}}}{0.25 \times M_{\text{ethanol}} + (1 - 0.25) \times M_{\text{water}}} \right) \times 100 ]