Question

Determine how much water should be mixed with 6 gallons of 55% solution of sulfuric acid to make a 20% solution of sulfuric acid?
Thanks

Answer 1

Answer:

10.5 gallons of water should be added.

Step-by-step explanation:

Let $C_{1} * V_{1} = C_{2} * V_{2}$, where $C_{i}$ are the concentration of the solution and $V_{i}$ are the volume of the solution for i=1,2.

Now, substituting values, we have,

55*6 = 20*$V_{2}$

i.e. $V_{2}$ = (6*55)/20

i.e. $V_{2}$ = 16.5

So, the final volume is 16.5 gallons.

Hence, we need to add 16.5 - 6 = 10.5 gallons of water.

Answer 2

Answer: 16.5 gallons of water should be mixed to make a 20% solution of $H_2SO_4$

Step-by-step explanation: We are given the concentration of the solution in percentage, it can be taken as the molarity of the solution.

Now, to calculate the volume of water required, we use the formula:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of the solution.

$M_2\text{ and }V_2$ are the final molarity and volume of the solution.

Given:

$M_1=55\%\\V_1=6gallons\\M_2=20\%\\V_2=?gallons$

Putting values in above equation, we get:

$55\times 6=20\times V_2\\V_2=16.5gallons$