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20 February, 06:08

A pharmacist wants to mix a 16% saline solution with a 26% saline solution to get 15 l of a 20% saline solution. how much of each solution should she use?

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  1. 20 February, 08:28
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    the pharmacist wants to mix 2 solutions to make 15 L of a solution with 20% percentage

    so if he mixes a volume of y of 16%, the quantity of solute added from the 16% solution is 16% x y = 0.16y

    and he mixes a volume of z of 26%, the quantity of solute added from 26 % solution is 26% x z = 0.26z

    the final solution has a solute quantity of 20 % x 15 L = 0.2 x 15 = 3

    quantities of solute added from both solutions is equal to the quantity of the final solution

    0.16y + 0.26z = 3 - --1)

    the 2 volumes added equals 15 L

    y + z = 15 - --2)

    then we have a set of simultaneos equations

    multiply 2nd equation by 0.16

    0.16y + 0.16z = 2.4 - -3)

    subtract 3rd equation from the 1st equation

    0.26 z - 0.16z = 0.6

    0.10 z = 0.6

    z = 0.6/0.10 = 6

    since z = 6

    then substitute z = 6 in 2nd equation

    y + z = 15

    y + 6 = 15

    y = 9

    the volume of 26% solution added is 6 L

    volume of 16 % solution added is 9 L
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