Question

You need a 35% alcohol solution. On hand, you have a 440 mL of a 15% alcohol mixture. You also have 90%alcohol mixture. How much of the 90% mixture will you need to add to obtain the desired solution?You will needmL of the 90% solutionto obtainmL of the desired 35% solution.

Answer 1

We need a 35% alcohol solution.

We have:

15% alcohol: 440 mL

90% alcohol: To be determinate (x)

If we add x mL of 90% alcohol, we will have: (440 + x) mL of solution. 35% of that is alcohol. But of those 440 mL and x mL, 15% and 90% is alcohol, respectively, so the amount of alcohol in the final solution is:

(15% of 440) + (90% of x) = (35% of (400 + x))

440*15/100 + 90*x/100 = 35*(440 + x)/100

We cancel the common denominator:

440*15 + 90*x = 35*(440 + x)

6600 + 90x = 15400 + 35x

Solving for x:

90x - 35x = 15400 - 6600

55x = 8800

x = 160

We need to add 160 mL of 90% alcohol. The desired 35% solution has (440 + 160) mL = 600 mL