Question

How many liters of water must be added to 50 liters of 70% ethyl alcohol to reduce it to a mixture of 40% ethyl alcohol concentration?

Answer 1

Step-by-step explanation:

Let x = amount (liters) of 70% added

then

70x + .40(50) = .50(x+50)

70x + 20 = .50x+25

20x + 20 = 25

20x = 5

x = 5/.20

x = 25 liters

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Answer 2

The answer is 37.5 liters of water.

GIVEN

Total volume of solution = 50 litres.

Percentage of ethyl alcohol = 70%

TO FIND

Liters of water that should be added .

SOLUTION

We can simply solve the above problem as follows;

Initial Volume of solution = 50l

Percentage of ethyl alcohol = 70%

Volume of ethyl alcohol = (70/100) × 50 = 35 litres.

Let the volume of water added = X litres.

New volume = 50 + X litres.

Volume of ethyl alcohol = 40% of (50+X)

Therefore,

35 = 40% Of (50 + X)

35 = 40/100(50+X)

35 = 0.4 (50 +X)

35 = 20 + 0.4X

0.4X = 35-20

0.4X = 15

X = 37.5l

Hence, The answer is 37.5 liters of water.

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