Question

A chemist has two solutions of HNO3. One has a 40% concentration and the other has a 25% concentration.

How many liters of each solution must be mixed to obtain 57 liters of a 26% solution?

Answer 1

Explanation:

Let +a+ = liters of 40% solution needed

Let +b+ = liters of 25% solution needed

+.4a+ = liters of HN03 in 40% solution

+.25b+ = liters of HN03 in 25% solution

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(1) +%28+.4a+%2B+.25b+%29+%2F+57+=+.26+

(2) +a+%2B+b+=+57+

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(1) +.4a+%2B+.25b+=+.26%2A57+

(1) +.4a+%2B+.25b+=+14.82+

(1) +40a+%2B+25b+=+1482+

Multiply both sides of (2) by +25+

and subtract (2) from (1)

(1) +40a+%2B+25b+=+1482+

(2) +-25a+-+25b+=+-1425+

+15a+=+57+

+a+=+3.8+

and, since

(2) +a+%2B+b+=+57+

(2) +3.8+%2B+b+=+57+

(2) +b+=+53.2+

3.8 liters of 40% solution are needed

53.2 liters of 25% solution are needed

check:

(1) +%28+.4a+%2B+.25b+%29+%2F+57+=+.26+

(1) +%28+.4%2A3.8+%2B+.25%2A53.2+%29+%2F+57+=+.26+

(1) +%28+1.52+%2B+13.3+%29+%2F+57+=+.26+

(1) +14.82+=+.26%2A57+

(1) +14.82+=+14.82+

think you should got the answer