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?

Question

Chemistry

Mclean

26 October, 15:00

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?

+1

Answers (1)

Know the Answer?

Gilbert Merritt · Answer 1

Answer a) is 2000 mL solution 1.

Answer b) is 40 mL solution 1 and 60 mL solution 2.

Answer c) isn't possible to produce a solution with 90% benzene and 10% toluene because is a percentage of benzene higher than the solution 1 and solution 2.

Explanation:

You can solve using an equation system base on the benzene percentage/ratio of each solution.

Answer a) X-> mL of solution 1

x*0.8+500*0.3 = (500 + x) * 0.7

x * 0.8+500*0.3=500*0.7 + x*0.7

x * (0.8-0.7) = 500 * (0.7-0.3)

X*0.1=200

x=2000mL

Answer b) X-> mL of solution 1

x*0.8 + (100-x) * 0.3 = (100) * 0.5

x * 0.8+100*0.3-x*0.3=50

x * (0.8-0.3) = 50-30

x * (0.5) = 20

x=40 mL

mL of solution 2 = (100-x) = 100-40=60 mL