Question

What volume (in L) of a 4.13 M lithium nitrate `LiNO_3` solution would be needed to make 195 mL of a 1.05 M solution by dilution?

Answer 1

`C_1V_1=C_2V_2`

`V_2~=*50*mL`

Explanation:

`"Concentration"` `=` `"Moles of solute"/"Volume of solution"`

Thus `"Moles"` `=` `"Volume "xx" concentration"`

So `C_1V_1=C_2V_2`, where `C="Concentration; V=Volume"`

And `V_1=(C_2V_2)/C_1` `=` `(1.05*mol*L^-1xx0.195*L)/(4.13*mol*L^-1)` `=` `??L`