Question

You need 960 mL of a 40% alcohol solution. On hand, you have a 20% alcohol mixture. How much of the 20% alcohol mixture and pure alcohol will you need to obtain the desired solution?

You will need
mL of the 20% solution
and
mL of pure alcohol.

Answer 1

the 50% concentration of the final solution is a result of mixing the 95% concentration with water, which is 0% concentration.

Multiply the concentration of each liquid times the # of ml you will use, add these and divide by the total ml to get 50%

X = the number of ml of the 95% alcohol mixture

Y = the number of ml of water

X + Y = 855

{.95·X + (0)(Y } / 855 = .50

Again, the 0 above represents the water which is 0% concentration

.95X / 855 = .50

.95X = 427.50

X = 450

plug 450 into the equation X + Y =855

450 + Y =855

Y = 405

Use 450 ml of the 95% solution and 405 ml of water