Question

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?

Answer 1

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")`.

Answer 2

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.

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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.