A news reporter claims that by mixing 3 bottles of 30% ethyl alcohol, one would be able to produce 90% ethyl alcohol. Is he correct? How many mL of pure ethyl alcohol must be added to 200 mL of 30% ethyl alcohol to produce 90% ethyl alcohol?

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Messiah Trevino · Accepted Answer

No, news reporter is not correct because by mixing up less concetrated we can not get high concetrated.Volume of alcohol in 200 ml solution  =  0.30  ×  2000  =  60   mlvolume of other liquid = 200-60=140 mlLet x ml of ethyl alcohol is mixed up to make 90% strongSo volume of solution   =  60  +  x  +  140    =  200  +  xNow,  0.90  x  (  200  +  x  )  =  60  +  x    ⇒  180  +  0.90  x  =  60  +  x    ⇒  180  −  60  =  x  −  0.90  x    ⇒  120  =  0.10  x    ⇒  x  =      120    0.10    =  1200   mlSo, 1200 ml of pure ethyl alcohol must be added to get given result.

A news reporter claims that by mixing 3 bottles of 30% ethyl alcohol, one would be able to produce 90% ethyl alcohol. Is he correct? How many mL of pure ethyl alcohol must be added to 200 mL of 30% ethyl alcohol to produce 90% ethyl alcohol?

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