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26 July, 22:24

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

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  1. 27 July, 00:17
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    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.
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