An alcohol-water solution is 45.0 percent alcohol. How much solution is required to obtain 150.0 mL of alcohol?

Sagot :

Answer:

46.4 mL

Explanation:

The trick here is to use the known volume by volume percent concentration,

v/v%

, to figure out how many milliliters of isopropanol, the solute, are present in your sample.

Now, the thing to remember about volume by volume percent concentrations is that they are meant to represent the number of milliliters of solute present for every

100 mL

of solution.

In your case, the solution contains

74.3

%

v/v isopropanol

=

74.3 mL

.

isopropanol in

.

100 mL

.

solution

As you know, solutions are homogeneous mixture, which implies that they have the same composition throughout.

This means that you can use solution's volume by volume percent concentration to calculate the number of milliliters of isopropanol present in

62.4 mL

of solution.

Set up the equation as

?

.

mL solute

62.4 mL solution

=

74.3 mL

.

solute

100 mL

.

solution

Rearrange and solve to find

?

=

62.4

mL solution

100

mL solution

74.3 mL solute

?

=

46.36 mL solute