Question

350 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 1

a) We can say that x mL of solution 1+500 mL of solution 2= 500+x mL of what we call solution 3. We can get the equation 0.8*x+0.3*500=0.7(500+x). Simplifying this we get 0.8x+150=0.7x+350. Subtract 0.7x from both sides to get 0.1x+150=350. Subtract 150 from both sides to get 0.1x=200 the multiply both sides by 10 to get x=2000. So 2000 mL of solution 1 is needed to produce a solution that is 70% benzene.
b) We can say that the amount of solution 1 needed is x and the amount of solution 2 needed is 100-x. So we can get the equation 0.8*x+0.3(100-x)=0.5*100.Simplifying this we get 0.8x+30-0.3x=50. Simplifying even further we get 0.5x+30=50. Then by subtracting 30 from both sides we get that 0.5x=20 then by multiplying both sides by 2 we get x=40.So to form a 100 mL solution that is 50% benzene and 50% toluene we need 40 mL of solution 1 and 60 ml of solution 2.
c)no solution