Question

What volume of 10.0 M solution is needed to prepare 4.00 L of 0.500 M solution?

Answer 1

0.200 L

Step-by-step explanation:

To understand these exercises, one has to know the variables we are working with, what we are looking for, and also what equation to use. The variables we have are the following:

`C_(1)` = 10.0 M

`V_(2)` = 4.00 L = 4000 mL

`C_(2)` = 0.500 M

The subscript 1 belongs to the first solution and the subscript 2 belongs to the second solution. C = Concentration and V = Volume

We have to find out the volume needed of the solution 1 to prepare solution 2, so:

`V_(1)` = ?

We know that we will have the same amount of mols in solution 1 and solution 2:

`# mols_(1) = # mols_(2)`

We can infere that mols = V x C (only in M), so:

`V_(1) * C_(1) = V_(2) * C_(2)` ⇒ We have to solve for V1 ⇒
`V_(1) = (V_(2) * C_(2) )/(C_(1) )`

`V_(1) = (4000 mL * 0.500 M)/(10 M)` ⇒
`V_(1)` = 200 mL = 0.200 L

The most correct answer is 0.200 L (Note: Always have in mind the significant figures, which in this case is three sig. figs. Don’t lose points in something preventable!)

Answer 2

The answer to your question is 0.2 l or 200 ml

Step-by-step explanation:

Data

Volume 1 = ?

Concentration 1 = 10 M

Volume 2 = 4 l

Concentration 2 = 0.50 M

Process

1.- To solve this problem use the dilution equation

Volume 1 x Concentration 1 = Volume 2 x Concentration 2

-Solve for Volume 1

Volume 1 = Volume 2 x Concentration 2 / Concentration 1

2.- Substitution

Volume 1 = 4 x 0.5 / 10

3.- Simplification

Volume 1 = 2 / 10

4.- Result

Volume = 0.2 l or 200 ml