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# Need Help!

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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?

Mar 12, 2019

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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?

Let x be the mL of solution 1   that must be added

x(.80)  + (500)(.30) =  (500 + x)(.70)

.80x + 150  =  350 +  .70x       rearrange as

.10x =  200     divide both sides by  .10

x = 2000 mL   Mar 12, 2019
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Mar 12, 2019
edited by Guest  Mar 12, 2019
#3
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Mar 12, 2019
edited by Guest  Mar 12, 2019
#4
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Solution 1 consists of 80% benzene and 20% toluene. Solution 2 consists of 30% benzene and 70% toluene.

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?

If the final solution is 50% of both substances, then 50 mL  must be benzene and 50 mL must be toulene

Let x be the amount of solution 1  and y be the amount  of solution 2....and we have

x(.80) + y(.30)  =  50    ⇒   8x + 3y  = 500          (1)

x(.20) + y(.70) =  50     ⇒  2x + 7y =  500 ⇒   -8x  - 28y  = -2000   (2)

Add (1)  and (2)  and we get that

-25y = -1500   divide both sides by  -25

y =  60 mL   of solution 2

So...we need 40 mL of solution 1

c)   This is impossible .....the greatest concentation of benzene would be 80%....in other words, if only solution 1 were present.....any mixing with solution 2 would dilute this.....so 90% benzene is  impossible   Mar 12, 2019
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Thank you so much I did not understand this problem at all before!

Mar 13, 2019