A #1.5*dm^3# volume of #0.30*mol*dm^-3# #NaCl(aq)# was mixed with a #2.5*dm^3# volume of #0.70*mol*dm^-3# #NaCl(aq)#. What is the final concentration of #NaCl(aq)#?

1 Answer
Oct 27, 2016

Answer:

#[NaCl(aq)]=0.55*mol*L^-1#.

Explanation:

#"Concentration"# #=# #"Moles of solute"/"Volume of solution"#.

We have to calculate the quantity of all the ions in solution, and then divide this by the total volume.

#"Moles of NaCl(i)"# #=# #1.50*dm^3xx0.30*mol*dm^-3=0.45*mol#

#"Moles of NaCl(ii)"# #=# #2.50*dm^3xx0.70*mol*dm^-3=1.75*mol#

And thus #[NaCl]# #=# #(0.45*mol+1.75*mol)/(4.0*dm^3)#

#=# #0.55*mol*dm^-3#.

Note that #1*dm^3# #=# #(10^-1m)^3# #=# #10^-3m^3=1L#. We have (reasonably) assumed the volumes of solution to be additive.