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?
+129850 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
+129850 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
