What volume of a #1.5*mol*L^-1# solution is required to deliver a #3.0*mol# quantity of solute?

2 Answers
anor277
Mar 23, 2017

#2.0*L# of solution are required.

Explanation:

Look at the problem dimensionally:

we know that #"Moles of solute"/"Volume of solution"=1.5*mol*L^-1#

So #"Moles of solute"-="Volume of solution"xx1.5*mol*L^-1#

OR #"Volume of solution"="Moles of solute"/(1.5*mol*L^-1)#

#=(3.0*cancel(mol))/(1.5*cancel(mol)*L^-1)=2*L#

We get an answer in #L# because #1/L^-1=1/(1/L)=L# as required.

In order to use these quotients, you have to remember the basic definition:

#"Concentration"# #=# #"Moles of solute (mol)"/"Volume of solution (L)"#; and thus #"concentration"# is commonly quoted with units of #mol*L^-1#.

Capisce?

Simon Moore
Mar 23, 2017

2 litres.

Explanation:

To actually work it out, use the simple relationship #M = n/V#

Where M is molarity (or concentration) in moles per #dm^3# (or moles per litre, which is numerically identical), n is the number of moles of solute, and V is the volume of solution in #dm^3# (or litres, which are numerically identical).

Then simply rearrange for V, so #V = n/M# and plug in the numbers:

#V = n/M = 3/1.5 = 2# litres

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