Question

A chemist is adding pure (100%) acid to a 12 liter solution that is 15% acid in order to increase it to 49% acid. How much pure acid must she add?
How many ounces of 35% ethanol must be added to a 23 ounce mixture that contains 26% ethanol in order to obtain a mixture of 30.86% ethanol?

How many liters of 30% acid must be added to 10 liters of 53.66% acid in order to obtain a mixture that is 43% acid?

Answer 1

Answer:

1) 8 liters of pure acid

2) 27 liters of 35% ethanol

3) 8.2 liters of 30% acid

Step-by-step explanation:

The inicial amount of acid is 15% of 12 L, and the inicial volume is 12L.

Adding 'x' L of 100% acid, we have the amount of acid equal to 12 * 0.15 + x * 1, and the volume of 12 + x. The concentration is the amount of acid over the volume, so we have that:

(12*0.15 + x) / (12 + x) = 49/100

(1.8 + x)*100 = (12 + x) * 49

180 + 100x = 588 + 49x

51x = 408

x = 8 liters of pure acid

Similar to the solution above, for the second question we have:

(23*0.26 + 0.35*x) / (23 + x) = 30.86/100

(5.98 + 0.35*x)*100 = (23 + x) * 30.86

598 + 35x = 709.78 + 30.86x

4.14x = 111.78

x = 27 liters of 35% ethanol

For the third question, we have:

(10*0.5366 + 0.3*x) / (10 + x) = 43/100

(5.366 + 0.3x)*100 = (10 + x)*43

536.6 + 30x = 430 + 43x

13x = 106.6

x = 8.2 liters of 30% acid