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 #"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.