# 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?

Jun 5, 2017

The amount of $\text{RbI}$ that was added is $8800$ $\text{g}$.

#### Explanation:

For this kind of questions, we use the formula:

$\left({\text{mol")/("volume "color(blue)("L"))="concentration " color(blue)("mol"xx"L}}^{-} 1\right)$

To calculate the mass of rubidium iodide ($R b I$), 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:

$\text{mol"="volume"xx"concentration}$
"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 $R b I$. 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!

$\text{mol"="volume"xx"concentration}$
"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 $R b I$ into mass units and done! We use the molar mass of $R b I$ to convert. We can calculate the molar mass from adding the atom masses $u$ or looking it up:

$\text{molar mass}$ $R b I = 212.37 \textcolor{b l u e}{{\text{ gram"xx"mol}}^{-} 1}$

$\text{mass "= "molar mass "xx" mol}$

"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 $\textcolor{red}{\text{g}}$.

Jun 5, 2017

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

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

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

Looking at the end of the question, we were given $\text{13.9 L}$ of a $\text{0.65 M}$ solution, which has:

$13.9 \cancel{\text{L" xx "0.65 mols"/cancel"L" = "9.035 mols RbI}}$ for every $\text{L}$ of solution.

The number of mols in the $\text{1 L}$ sample of the solution and the number of mols in the $\text{13.9 L}$ 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 $\text{4.6 L}$ solution (from which the $\text{1 L}$ was taken at fixed concentration!), we have:

$\text{9.035 mols RbI" xx (4.6 cancel"L")/cancel"1 L}$

$=$ $\text{41.561 mols RbI}$ in that $\text{4.6 L}$ of solution.

Therefore, you had a mass of:

$\textcolor{b l u e}{{m}_{R b I}} = 41.561 \cancel{\text{mols RbI" xx "212.3723 g"/cancel"1 mol RbI}}$

$=$ $\text{8826.41 g}$

$\to$ $\textcolor{b l u e}{\text{8800 g}}$ to two sig figs

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

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

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