How many milliliters of 11.5 M HCl(aq) are needed to prepare 305.0 mL of 1.00 M HCl(aq)?

1 Answer
Sep 1, 2016

Answer:

Approx. #27*mL#

Explanation:

#C_1V_1=C_2V_2#, where #C_i# is the concentration of a component in #mol*L^-1#, and it follows that the product #C_iV_i# gives an answer in the number of moles #(CV="concentration"(mol*L^-1)xx"volume"(L)="moles")#

And thus #"moles"# #=# #C_1V_1# #=# #C_2V_2#.

We simply solve for #V_1=(C_2V_2)/C_1#

#=(305.0xx10^-3Lxx1.00*cancel(mol)*cancel(L^-1))/(11.5*cancel(mol)*cancel(L^-1)# #=# #??L#