Chemical Mixture Problems Type B In the chemical mixture problems worked thus far we have mixed two solutions of different percentage concentrations to get a mixture that has a different percentage concentration from either of the original solutions It is interesting to note that the percen ID: 572271
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Slide1
Lesson 61:
Chemical Mixture Problems, Type BSlide2
In the chemical mixture problems worked thus far, we have mixed two solutions of different
percentage
concentrations to get a mixture that has a different percentage concentration from either of the original solutions. It is interesting to note that the percentage concentration of the final mixture must fall between the percentage concentrations of the two solutions used. Slide3
For example, if we pour in some 15 % iodine solution and dump in some 60% iodine solution, we can get a final mixture whose concentration is somewhere between 15% iodine and 60% iodine. It is impossible to mix these solutions and get a final concentration greater
than
60% iodine or less than 15% iodine. Slide4
For want of a better name, we will call problems like the one before “type A chemical mixture problems”. Now we will look at another kind of chemical mixture problem. We will call these “type B problems”. Slide5
In a type B problem we will begin with a given mixture and add to it or remove something from it. As in type A problems, the equation will make a statement about one of the components of the mixture. Slide6
Example:
How much water must be evaporated from 100 gallons of a 10% brine solution to get a 40% brine solution?Slide7
Answer:
Water one – water out = water final
100 – E = (100 – E) 0.9(100) – E = 0.6(100 – E)
E = 75
Salt one – salt out = salt final
100 – E = (100 – E) 0.1(100) – 0(E) = 0.4(100 – E)
E = 75
We must evaporate 75 gallons of water to get a mixture that is 40% salt. Slide8
Example:
When Frank and Mark finished milking, they found that they had 900 pounds of milk that was 2 percent butterfat. How much butterfat did they have to add to raise the butterfat content to 8 percent? (Whole milk is a mixture of skim milk and butterfat.)Slide9
Answer:
Butterfat one + Butterfat added = Butterfat final
(900) + (P) = (900 + P)
0.02(900) + 1(P) = 0.08(900 + P)
P = 58.7 pounds
Thus, 58.7 pounds of butterfat should be added to get a mixture that is 8 % butterfatSlide10
Example:
Virginia and Campbell had 100 kilograms of a 20% glycol solution. How much of a 40% glycol solution should they add to get a solution that is 35% glycol?Slide11
Answer:
g
lycol One + glycol added = glycol final
(100) + P = (100 + P)
0.2(100) + 0.4(P) = 0.35(100 + P)
P = 300
Thus, if they pour in 300 kg of a 40% glycol mixture, the result will be a mixture that is 35% glycol.Slide12
HW: Lesson 61 #1-30