PLEASE ANSWER How many liters of a 35% salt solution must be mixed with 20 liters of 70% salt solution to obtain a solution that is 45% salt?

The volume of the 35% solution is 50 L

Known parameter;

The percentage concentration of the salt solution with unknown volume = 35%

The percentage concentration of the 20 liters salt solution = 70%

The percentage concentration of the resulting solution = 45%

Unknown parameter;

The volume of the 35% salt solution

Strategy;

Let x represent the volume of the 35% salt solution that must be mixed  to obtain the 45% salt solution

Balance the masses of the salts before and after mixing

The mass of the salts before mixing = 70% × 20 L + 35% × x

The mass of the salt solution mixture after mixing = 45% × (20 L + x)

By conservation principles, we get;

70% × 20 L + 35% × x = 45% × (20 L + x)

0.7 × 20 L + 0.35 × x = 0.45 × (20 L + x)

14 L + 0.35·x = 9 L + 0.45·x

14 L - 9 L = 0.45·x - 0.35·x = 0.1·x

5 L = 0.1·x

x = 5 L/(0.1) = 5 L × 10 = 50 L

The volume of the 35% solution that must be mixed with the 70% salt solution x =  50 L

Learn more about calculations of concentration of mixtures here;

link

50 liters

Step-by-step explanation:

Amount of salt in the 35% solution + amount of salt in the 70% solution = amount of salt in the 45% solution

0.35 x + 0.70 (20) = 0.45 (x + 20)

0.35 x + 14 = 0.45 x + 9

5 = 0.1 x

x = 50

20

step-by-step explanation: