How much pure antifreeze must be added to 12 L of a 40% solution to obtain a 60% solution?

1 Answer
Jun 29, 2017

The answer is 6 L.

Explanation:

First we need to determine the amount of antifreeze in the 12 L solution.

0.4 = x/120.4=x12 rArr x = 12*0.4 = 4.8 Lx=120.4=4.8L

Then using the formula below:

0.60.6 = (4.8 + x)/(12+x)4.8+x12+x

0.6 (12+x)0.6(12+x) = (4.8 + x)(4.8+x)

7.2 + 0.6 x7.2+0.6x = 4.8 + x4.8+x

7.2 - 4.87.24.8 = x - 0.6 xx0.6x

2.4 = 0.4 x2.4=0.4x

x = 6 Lx=6L