Question

What volume of `12*mol*L^-1` `NaOH` is required to make a `1*L` volume of `4*mol*L^-1` concentration?

Answer 1

We use the relationship `C_1V_1=C_2V_2` to get `V_1(NaOH)=1*L`

Explanation:

The product `CxxV` gives us an answer in `"moles"`.

And thus we need `3*Lxx4*mol*L^-1=12*mol` `NaOH`. And thus we take a `1*L` volume of `12*mol*L^-1` `NaOH`, and dilute this to `3*L`.

So `"concentration"=(1*Lxx12*mol*L^-1)/(3*L)=4*mol*L^-1`.....as required.