How many mL of 10.0 M HCI are needed to prepare 500. mL of 2.00 M HCl?

1 Answer
Jan 30, 2017

Answer:

Approx. #100*mL#

Explanation:

The product, #"concentration"xx"volume"=mol*L^-1xxL#, clearly has units of #mol#, i.e. #mol*cancel(L^-1)xxcancelL=mol#.

And thus #C_1V_1=C_2V_2#, and given the equality we can use units of #mL# for volume.

#V_1=(C_2V_2)/C_1# #=# #(500*mLxx2.00*mol*L^-1)/(10.0*mol*L^-1)#

#=??mL#

AND YOU ADD THE CONCENTRATED ACID TO THE WATER NEVER THE REVERSE. WHY NOT?