Question

What volume of `0.315*mol*L^-1` `NaOH` is required to deliver `6.22*g` of `NaOH`?

Answer 1

Approx. half a litre of the given solution is required.........

Explanation:

We require `(6.22*g)/(40.00*g*mol^-1)=0.0156*mol` WITH RESPECT TO `NaOH`.

Now, by definition, `"Concentration"="Moles"/"Volume"`......

And thus `"Volume"="Moles"/"Concentration"=(0.0156*mol)/(0.315*mol*L^-1)`

`=0.494*L`.