# Question #c8e75

##### 1 Answer

#### Answer:

Here's how you could do that.

#### Explanation:

Judging from the amount of information given, I will have to make some assumptions in order to be able to provide an answer.

SInce no mention was made about what type of percent concetration you have, I'll assume that its *weight by weight*, or percent concentration by mass,

Now, a *sodium lactate*, *for every*

Let's assume that you set the volume of the **one liter**. This implies that the solution must contain

#C = n/V implies n = C * V#

#n = "0.45 M" * "1 L" = "0.45 moles"#

of sodium lactate.

Use the compound's molar mass to determine how many grams would contain this many moles

#0.45color(red)(cancel(color(black)("moles"))) * "112.06 g"/(1color(red)(cancel(color(black)("mole")))) = "50.43 g"#

Now use the sodium lactate's known percent concentration to find how many grams of

#50.43color(red)(cancel(color(black)("g sodium lactate"))) * "100 g solution"/(50color(red)(cancel(color(black)("g sodium lactte")))) = "100.86 g solution"#

So, to prepare a **enough water** to make the volume of the final solution equal to **one liter**.

As practice, you can redo the calculations for different volumes of the final solution.

A different volume of the final solution would affect the number of moles of sodium lctate it must contain, which in turn will affect the mass of the