# Question c8e75

Oct 23, 2015

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, $\text{w/w}$.

Now, a 50%"w/w" percent concentration tells you that you hve $\text{50 g}$ of solute, which in your case is sodium lactate, $\text{C"_3"H"_5"O"_3"Na}$, for every $\text{100 g}$ of solution.

Let's assume that you set the volume of the $\text{0.45-M}$ solution to one liter. This implies that the solution must contain

$C = \frac{n}{V} \implies n = C \cdot V$

$n = \text{0.45 M" * "1 L" = "0.45 moles}$

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%"w/w" solution would contain this many grams of sodium lactate

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 $\text{1-L}$, $\text{0.45 M}$ sodium lacte solution, take approximately $\text{101 g}$ of the 50%"w/w" sodium lactate solution and add 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 50%# solution that you must use.