When 0.50 liter of 12 M solution is diluted to 1.0 liter, what is the molarity of the new solution?

1 Answer
Jan 5, 2017

Answer:

The new concentration is HALF that of the original.

Explanation:

#"Concentration (C)"# #=# #"moles of solute (n)"/"volume of solution (V)"#.

Since #C=n/V#, #n=CV#.

And thus #n_"initial"# #=# #0.50*cancelLxx12*mol*cancel(L^-1)=6*mol#

But #V_"final"# #=# #1.0*L#.

So #"concentration"#

#=# #(0.50*cancelLxx12*mol*cancel(L^-1))/(1*L)=??mol*L^-1#

You use the relationships, #C=n/V#, #V=n/C#, and #n=CV# continually in a laboratory.

Concentrated hydrochloric acid is supplied as a #10*mol*L^-1# solution in water. If I have a #2.5*L# bottle of conc. acid, how many litres of #1.0*mol*L^-1# can I prepare?

IMPORTANT: WE WOULD ALWAYS ADD CONC ACID TO WATER AND NEVER THE REVERSE!!