vhalalfonso
Answered

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How much water must be added to 220 gallon of mixture which is 80% alcohol to reduce it to 70% alcohol?

Sagot :

KanataHoshijima31

Answer:

31.43 gal.

Step-by-step explanation:

This is a concentration problem where we are given an initial solution with a certain concentration and we need to add something to reduce or increase the concentration.

The first step is to determine which part of the mixture is alcohol and which part is water. To work with the given, we need their exact amounts, not just their fractional part. To find this, we need to multiply 80% to 220 to find the alcohol and 20% to 220 to find the initial water.

220 x 80% = 220 x 0.8 = 176 gallons of alcohol

220 x 20% = 220 x 0.2 = 44 gallons of water

Now, to reduce the concentration, we need to add water. Let's understand first how we get the concentration. The formula for this is C = S / T where C is the concentration rate, S is the solute, and T is the total amount. We can utilize this formula and re-arrange it in our favor. If C=S/T, then T=S/C. We already know S, which is 176, and C which is 70%.

Let's now substitute and solve for T.

T = 176 gal / 70%

T = 176 gal / 0.7

T = approx. 251.43 gal

Now, initially we had 220 gallons of mixture. To find the amount added to this, we subtract 220 from 251.43. The answer to this operation is 31.43 gal.

Therefore, we need to add 31.43 gallons of water to a 220-gallon mixture of 80% alcohol solution to reduce the concentration to 70%

Hope this helps!

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