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

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Answer 1 · 2016-10-27T20:38:39

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

#"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.

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)#?