# 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

#### Answer:

The amount of

#### Explanation:

For this kind of questions, we use the formula:

To calculate the mass of rubidium iodide (

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

So in our 1 L solution, we had 9.035 mol

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!

Now we only have to convert mol

Which, when rounded to two significant figures (the lowest number given in the problem), is

Here is another way to do this problem using the extensive property of countable numbers. I also get

**OVERVIEW:**

Think critically through the problem... What did we do? We started with some starting mols of

That was then diluted from

Looking at the end of the question, we were given

#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

*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 ** 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 *mols are extensive*.

That was then diluted *the same number of mols* of