meli25
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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 :

olemakpadu
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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