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)   (Show Source): You can put this solution on YOUR website!
(a)
Let = ml of solution 1 to be added
= ml of benzene in solution 1
ml of benzene in solution 2
----------------------------------------





2000 ml ( 2 l ) of solution 1 must be added
check:





OK
----------
(b)
Let = ml of solution 1 needed
Let = ml of solution 2 needed
----------
(1)
(2)
---------------------------
(2)
(2)
(2)
Multiply both sides of (1) by
and subtract (1) from (2)
(2)
(1)


and
(1)

40 ml of solution 1 is needed
60 ml of solution 2 is needed
------------
(c)
No. You can't get a greater % of benzene than
either of the 2 solutions have which is
80% and 30%


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