Question

Question #30f4f

Answer 1

We WOULD normally use `C_1V_1=C_2V_2`, so.........`C_2~=0.02*mol*L^-1`

Explanation:

The product `C_1V_1` has units `mol*L^-1xxL`, i.e. you get an answer in `mol` as required. Here, we want the `"concentration"`, that is given by the quotient `"moles"/"volume"`..........

And so here we use this quotient.................

`C_1V_1=C_2V_2`, `C_2=(C_1V_1)/V_2`,

`C_2=(50.0*mLxx10^-3*L*mL^-1xx0.195*mol*L^-1)/(5.00xx10^2*mLxx10^-3*L*mL^-1)`

`=1/10C_1`

And this makes sense given that we have diluted the volume `"tenfold"`.