You are making a saline solution in a science class. One hundred milliliters of a 47.21% saline solution is obtained by mixing a 35% saline solution with a 68% saline solution.
a) How much 68% saline solution must you use? Answer in units of milliliters.
Answer (2)
We would use mixing rule here. Let C = Concentration V = Volume. Mix 1 + Mix 2 = Final Mix. C1V1 + C2V2 = CV
C1 = 35% = 0.35, V1 = Let it be x. C2 = 68% = 0.68, V2 = (100 - x). Since total volume = 100 ml. C = 47.21% = 0.4721, V = 100 ml. Applying equation:
0.35* x + 0.68*(100- x) = 100*(0.4721). 0.35x + 68 - 0.68x = 47.21 0.35x - 0.68x = 47.21 - 68 -0.33x = -20.79. Divide both sides by - 0.33 x = -20.79/-0.33 x = 63. Recall that x= 63, was for the 35% salinity.
For 68% salinity = (100-x) = (100 - 63) = 37 mL
Our answer is 37 mL for the 68% salinity.
To create a 100 mL saline solution with a concentration of 47.21%, you need approximately 37 mL of the 68% saline solution. The remaining volume will be filled with the 35% saline solution, which will be about 63 mL. This is calculated using the mixing rule for concentrations and volumes.
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