Mixture A is 10% ethanol by volume and mixture B is 30% ethanol by volume. How many milliliters of mixture B
Answers
Given : Mixture A is 10% ethanol by volume and
mixture B is 30% ethanol by volume.
To Find : How many milliliters of mixture B must be added to 200 milliliters
of mixture A to create a mixture that is 25% ethanol by volume
Solution:
200 milliliters of mixture A
Mixture A is 10% ethanol by volume
=> ethanol = (10/100) 200 = 20 ml
Let say 100M ml mixture B is mixed
mixture B is 30% ethanol by volume.
=> ethanol = (30/100)100M = 30M
Ethanol = 30M + 20 ml
Total Volume = 200 + 100M ml
25% ethanol by volume = (25/100)(200 + 100M)
= 50 + 25M ml
30M + 20 = 50 + 25M
=> 5M = 30
=> M = 6
=> 100M = 600
Hence 600 ml of mixture B must be added
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Solution:
200 milliliters of mixture A
Mixture A is 10% ethanol by volume
=> ethanol = (10/100) 200 = 20 ml
Let say 100M ml mixture B is mixed
mixture B is 30% ethanol by volume.
=> ethanol = (30/100)100M = 30M
Ethanol = 30M + 20 ml
Total Volume = 200 + 100M ml
25% ethanol by volume = (25/100)(200 + 100M)
= 50 + 25M ml
30M + 20 = 50 + 25M
=> 5M = 30
=> M = 6
=> 100M = 600