Question

What volume of ethanol and water must be mixed together to prepare 250ml of 60% volume by volume solution of alcohol.

It's urgent

Answer 1

Hey dear !!

So here's the solution of the given question :-

Let the volume of alcohol taken = x mL.

Volume percent = Volume of alcohol/ volume of solution* 100.

Or 60= x/ 250 * 100 or X (i.e. volume of ethanol mixed).

= 60*250/100

= 150 mL.

Ꭲᕼᗴᖇᗴᖴᝪᖇᗴ ᐯᝪᏞᑌᗰᗴ ᝪᖴ ᗯᗩᎢᗴᖇ ᗰᏆ᙭ᗴᗞ = 250- 150 = 100ᗰᏞ

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Answer 2

Answer:

The volume of ethanol, 150ml and volume of water, 100ml  must be mixed  to prepare 250ml of 60% v/v solution of alcohol.

Explanation:

Given: the volume by volume percentage of solution = 60%

The total volume of the solution = 250ml

Consider the volume of ethanol in the solution = x

We can find the  volume by volume percentage, from volume of the solute divided by volume of the solution.

%v/v = [ volume of solute (ethanol)/volume of solution]×100

$60=\frac{x}{250}*100$

$x=\frac{15000}{100}$

$x=150ml$

Therefore the volume of ethanol required $=150ml$

The volume of water required $=250-100=150ml$