A certain amount of rubidium iodide was dissolved in "4.6 L". A "1-L" aliquot was extracted an diluted to "13.9 L" to form a "0.65 M" solution in water. What mass of rubidium iodide was dissolved into the original solution?

2 Answers
Jun 5, 2017

The amount of "RbI" that was added is 8800 "g".

Explanation:

For this kind of questions, we use the formula:

("mol")/("volume "color(blue)("L"))="concentration " color(blue)("mol"xx"L"^-1)

To calculate the mass of rubidium iodide (RbI), we need to calculate back to the original solution. We start therefore at the end of the question and work back.

We have a 13.9 L solution with a concentration of 0.65 M, using the formula above, rewritten:

"mol"="volume"xx"concentration"
"mol"=13.9 color(red)(cancel(color(blue)("L"))) xx0.65 (color(blue)("mol"))/(color(red)cancel(color(blue)("L")))=9.035 color(blue)(" mol")

So in our 1 L solution, we had 9.035 mol RbI. Using the same formula again but rewritten, we can calculate the concentration of that 1 L solution:

"concentration" =("mol")/("volume")

"concentration"=(9.035 color(blue)(" mol"))/(1color(blue)("L"))=9.035 color(blue)(("mol")/("L"))

Since we took 1 L from the original solution, the 9.035 mol/L is the concentration of the original solution. We had 4.6 L of the original solution and we need the amount of mol, therefore, you guessed it, we use again the same formula!

"mol"="volume"xx"concentration"
"mol"=4.6 color(red)(cancel(color(blue)("L"))) xx9.035 (color(blue)("mol"))/(color(red)cancel(color(blue)("L")))=41.561 color(blue)(" mol")

Now we only have to convert mol RbI into mass units and done! We use the molar mass of RbI to convert. We can calculate the molar mass from adding the atom masses u or looking it up:

"molar mass" RbI =212.37 color(blue)(" gram"xx"mol"^-1)

"mass "= "molar mass "xx" mol"

"mass "=212.37 ("gram")/color(red)(cancel(color(blue)("mol")))xx41.561 color(red)(cancel(color(blue)("mol")))=8826.31 color(blue)("gram")

Which, when rounded to two significant figures (the lowest number given in the problem), is color(red)(8800 color(red)("g").

Jun 5, 2017

Here is another way to do this problem using the extensive property of countable numbers. I also get "8800 g" (two sig figs).

OVERVIEW:
Think critically through the problem... What did we do? We started with some starting mols of "RbI" dissolved in "4.6 L", and then took "1 L" of that, which contained a smaller chunk of the original mols.

That was then diluted from "1 L" to "13.9 L" to form the "0.65 M" solution, which contains the same number of mols as there were in the "1 L" solution.


Looking at the end of the question, we were given "13.9 L" of a "0.65 M" solution, which has:

13.9 cancel"L" xx "0.65 mols"/cancel"L" = "9.035 mols RbI" for every "L" of solution.

The number of mols in the "1 L" sample of the solution and the number of mols in the "13.9 L" sample are the same quantity, since just water was added to the same mols as before.

Mols are an extensive quantity, just like mass. If you have a greater volume of a certain concentration, you have more mols of it.

Therefore, in "4.6 L" solution (from which the "1 L" was taken at fixed concentration!), we have:

"9.035 mols RbI" xx (4.6 cancel"L")/cancel"1 L"

= "41.561 mols RbI" in that "4.6 L" of solution.

Therefore, you had a mass of:

color(blue)(m_(RbI)) = 41.561 cancel"mols RbI" xx "212.3723 g"/cancel"1 mol RbI"

= "8826.41 g"

-> color(blue)("8800 g") to two sig figs

Now think back through the problem... What did we do?

We started with some "41.561 mol"s of "RbI" dissolved in "4.6 L", and then took "1 L" of that, which contained "9.035 mols RbI" because mols are extensive.

That was then diluted 13.9-fold to form the "0.65 M" solution, which contains the same number of mols of "RbI" as in the "1 L" sample.