Question

How many liters of pure water should be mixed with a 18-L solution of 80% acid to produce a mixture that is 90% water?

Answer 1

126 liters of pure water should be added

Solution:

Let "x" be the liters of pure water

Then, 18 + x is the liters of final solution

18 Liter solution of 80% acid, which means 100 - 80 = 20 % of water is used

Pure water is 100%

Then according to question, we can say,

"x" liters of pure water 100 % is mixed with a 18 liter solution of 20 % water to produce a mixture (18 + x) liter that is 90 % water

Thus we frame a equation as:

Equation:

water + water = water

$x \times 100\% + 18 \times 20 \% = (18+x) \times 90\%$

$x \times \frac{100}{100} + 18 \times \frac{20}{100} = (18+x) \times \frac{90}{100}\\\\Simplify\ the\ above\ equation\\\\x + 0.2 \times 18 = (18+x) \times 0.9\\\\x + 3.6 = 16.2 + 0.9x\\\\0.1x = 12.6\\\\Divide\ both\ sides\ of\ equation\ by\ 0.1\\\\x = 126$

Thus 126 liters of pure water should be added