dimmak
A chemistry student is working with solution M, which has a 20% calcium chloride concentration. He needs to remove three-fourths of this solution and then add a new solution, solution N, to create a third solution, solution P, with double the volume of solution M. What must the calcium chloride concentration of solution N be to create a calcium chloride concentration of 37.5% in solution P?
a) 32%
b) 35%
c) 40%
d) 44%
e) 55%
We can let amount of solution M = 20 liters, so it has 0.2 x 20 = 4 liters of calcium chloride. We also remove ¾ x 20 = 15 liters of M, which includes ¾ x 4 = 3 liters of calcium chloride. Thus, we have 5 liters of M left (which includes 1 liter of calcium chloride), and we have to add 35 liters of solution N to make 40 liters of solution P (recall that solution P is double the amount of solution M).
We can x = the percent of calcium chloride in solution N and create the equation:
(1 + x/100 * 35)/40 = 0.375
1 + 35x/100 = 15
35x/100 = 14
35x = 1400
x = 40
Alternate Solution:
Assume we start with 4 liters of 20% solution. We will remove ¾ of it, which is 3 liters of 20% solution. Now we add 7 liters of unknown concentration x, and the result is 8 liters of 37.5% solution. Note that we need to end with twice the volume of what we started with, so we end with 8 liters of solution. We can create the following equation:
4(0.20) - 3(0.20) + 7x = 8(0.375)
0.8 - 0.6 + 7x = 3
7x = 2.8
x = 0.40, or 40%
Answer: C