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Emmett Swearson Term project This problem for the term project is a mixing solutions problem. It involves mixing two different solutions each containing different concentrations of pure drug within them in order to create a new solution with a percent strength between the percent strengths of the two original solutions that were mixed together. We are given a 32.5% solution and a 6.5% solution. Our goal is to create a 1 liter solution with a strength of 9.5% using a mixture of the 32.5% and 6.5% solutions. 1. First, we interpret the strength of the 32.5% solution. To interpret the strength we simply put the percent strength in the numerator as grams. The denominator of the fraction in interpreting percent strength is 100 mL. 32.5% 32.5 g 100mL 2. To interpret the strength of 6.5% we do the same thing that we did in problem #1. We put the percent strength in the numerator as grams while we put 100 mL in the denominator. 6.5% 6.5 g 100mL 3. To interpret the strength of a 9.5% we do the same thing that we did in the previous two questions. We take 9.5% and put it in the numerator as grams. We put 100 mL in the denominator of the fraction. 9.5% 9.5 g 100mL 4. In order to find the amount of 32.5% and 6.5% solution to mix together to get 1L of 9.5% we need to use Shane’s formula for mixing solutions to find the volume of 32.5% solution needed in the 1L 9.5% solution, which is our goal. F= 1000 (0.095-0.065) = 115.4 mL of 32.5% solution (0.325-0.065) Once we have figured out the amount of stronger solution we can figure out the volume of weaker strength solution need by subtracting the volume needed of the strongest strength solution from the volume of the 1L goal solution. Volume of 6.5% solution need= 1000mL- 115.4mL =884.6mL of 6.5% solution. Now that we have done the arithmetic to find out the volume of each solution needed for the intended goal we can mix them. To create a 1 Liter 9.5% solution we need to mix 884.6 mL of 6.5% solution with 115.4 mL of 32.5% solution. 5. We used Shane’s formula to figure out how many mL of 32.5% solution we needed in order to create our goal of 1000 mL of a 9.5% solution. Our result was that we need 115.4 mL of 32.5% solution. 6. In order to find how many grams of pure drug are in the 115.4 mL of 32.5% solution we multiply the volume (115.4 mL) by the decimal form of 32.5%. To find the decimal form of 32.5% we divide it by 100. 32.5/100= 0.325 Next, multiply 115.4 mL by 0.325 to find the amount of pure drug in the solution. 115.4 x 0.325= 37.5 g There is 37.5 grams of pure drug in the volume of needed 32.5% solution. 7. In order to find the volume of 6.5% solution needed in the intended 1 L, 9.5% solution we have to subtract the volume of 32.5% solution needed from the goal of 1000 mL to get the difference which is the answer for the volume of 6.5% solution needed to achieve our intended solution. So we subtract 115.4 mL from 1000 mL to find the volume of 6.5% solution needed. 1000-115.4= 884.6 We need 884.6 mL of 6.5% solution. 8. To find the amount of pure drug in the 884.6 mL of 6.5% solution, we multiply the volume of 6.5% solution by the decimal form of 6.5%. To find the decimal form of the percent we need to divide 6.5% by 100. 6.5%/100=0.065 in decimal form. Next, multiply 884.6 mL by 0.065 884.6 x 0.065= 57.5 g So, there is 57.5 grams of pure drug in the needed 884.6 mL of 6.5% solution needed to create a 1000 mL, 9.5% solution. 9. To calculate the sum in grams of pure drug in both 32.5% solution and 6.5% we add together our answer from questions #6 and #8. So, we add 37.5 to 57.5 to get the total amount of pure drug in the final solution. 37.5+57.5= 95 So, there is 95 grams of pure drug in our intended 1000 mL of 9.5% solution. 10. To directly compute the amount of pure drug within a 9.5% 1 liter solution we simply convert liters to milliliters and multiply it by the decimal form of 9.5%. First lets convert liters to millileters by moving the decimal 3 places to the left converting 1 L to 1000 mL. 1L 1000 mL Next we find the decimal form of 9.5%. To do this we divide 9.5% by 100. 9.5%/100=0.095 Finally, to directly compute the amount of pure drug within the 1 L 9.5% solution we multiply 1000 mL by 0.095. 1000 x 0.095= 95 g Results from our calculation show that there is 95 grams of pure drug in a 1 L 9.5% solution. Reflection The problem for this semesters math project was a mixing solutions problem. We took two solutions, each with different strengths and mixed them together to create a new solution with a specific strength and volume. After we correctly calculated the volume of each concentrated solution, we mix them together to get our final solution. If we correctly solved the equation, questions nine and ten will be equal to each other. In question nine we are asked to find the amount of pure drug (in grams) we added from each different solution to make the final solution. In question ten, we are asked to find the amount of pure drug (in grams) in the final solution. If we add the amount of pure drug from the two solutions we used to make the final solution are added together and equal the amount of pure drug we should have in our final solution then we have done our arithmetic correctly and we have correctly solved the problem. I learned from this project a new way of checking my answers by adding the amounts of drugs together then cross referencing them to the amount of pure drug in my intended solution. Medical math is more than just a passing grade. While I personally am not going into the medical field, this class has developed my math skills tremendously. I feel like I can complete problems in my head much faster than I could before and I feel much more confident in my math skills. Group projects in math class can be helpful. Sometimes having someone else besides the teacher explain how to correctly complete problems can make more sense to me. Sometimes if I don’t understand how to do a certain problem or get stuck I will ask another student and often times figure out the problem with just a little help. The only downside to group projects is that sometimes on group member will get left behind and be to afraid to ask the group questions so that group member can fully understand how to complete the problem. If one group member gets left behind it completely ruins the whole point of group work. I have really learned a lot from this medical math class and feel like my math skills are much stronger and efficient.