Question

How many ml of ethanol should be adding in 250ml of water which obtain a solution containing 80% of volume/volume solvent?

Answer 1

Answer:

200 ml

Explanation:

Let the ml of ethanol be added to water be x.

Now, volume by volume percent = $\sf \frac{volume \ of \ solute}{volume \ of \ solvent} \times 100$

So, given volume percent = 80%

Volume of solvent = 250 ml

Volume of solute = x

$\implies \frac{x}{250} \times 100 = 80 \\\\ \implies x = 80 \times \frac{250}{100}\\\\\implies x = 200 ml$

Hence, 200 ml of ethanol should be added to 250 ml of water to obtain a solution containing 80% of volume by volume solvent.

Answer 2

$Volume \: Percentage \: = \: \frac{volume \: of \: solute}{volume \: of \: solvent} \times 100$

Given volume percentage = 80

Volume of solvent = 250 ml

Let the volume of solute be = x

$80 = \frac{x}{250} \times 100 \\ \\ 80 \times 5 = 2x \\ \\ 200 = x$

The required value of x is 200 ml.

So the volume of the solvent to be added = 200 ml.