SOLUTION: Solution 1 consists of 80% benzene and 20% toluene. Solution 2 consists of 30% benzene and 70% toluene. a) How many mL of Solution 1 must be added to 500 mL of Solution 2 in or

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Question 763409: Solution 1 consists of 80% benzene and 20% toluene. Solution 2 consists of 30% benzene and 70% toluene.
a) How many mL of Solution 1 must be added to 500 mL of Solution 2 in order to produce a solution that is 70% benzene?
b) How many mL of Solution 1 and how many mL of Solution 2 must be combined to form a 100 mL solution that is 50% benzene and 50% toluene?
c) Is there a combination of Solution 1 and Solution 2 that is 90% benzene and 10% toluene?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(a)
Let +a+ = ml of solution 1 to be added
+.8a+ = ml of benzene in solution 1
+.3%2A500+=+150+ ml of benzene in solution 2
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+%28+.8a+%2B+150+%29+%2F+%28+a+%2B+500+%29+=+.7+
+.8a+%2B+150+=+.7%2A%28+a+%2B+500+%29+
+.8a+%2B+150+=+.7a+%2B+350+
+.1a+=+200+
+a+=+2000+
2000 ml ( 2 l ) of solution 1 must be added
check:
+%28+.8a+%2B+150+%29+%2F+%28+a+%2B+500+%29+=+.7+
+%28+.8%2A2000+%2B+150+%29+%2F+%28+2000+%2B+500+%29+=+.7+
+%28+1600+%2B+150+%29+%2F+2500+=+.7+
+1750+=+.7%2A2500+
+1750+=+1750+
OK
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(b)
Let +a+ = ml of solution 1 needed
Let +b+ = ml of solution 2 needed
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(1) +a+%2B+b+=+100+
(2) +%28+.8a+%2B+.3b+%29+%2F+100+=+.5+
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(2) +.8a+%2B+.3b+=+.5%2A100+
(2) +.8a+%2B+.3b+=+50+
(2) +8a+%2B+3b+=+500+
Multiply both sides of (1) by +3+
and subtract (1) from (2)
(2) +8a+%2B+3b+=+500+
(1) +-3a+-+3b+=+-300+
+5a+=+200+
+a+=+40+
and
(1) +a+%2B+b+=+100+
+b+=+60+
40 ml of solution 1 is needed
60 ml of solution 2 is needed
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(c)
No. You can't get a greater % of benzene than
either of the 2 solutions have which is
80% and 30%