Question

If water is added to 100.0 mL of a 0.150 M sodium hydroxide solution unit the final volume is 150.0 mL, what will the molarity of the diluted solution be?

Answer 1

The end concentration will be...........`0.100*mol*L^-1`

Explanation:

`"Concentration"="Moles of solute"/"Volume of solution"`, i.e. `C=n/V`.

So `n=CV`.

Since here `n` is constant:

`C_1V_1=C_2V_2`

And `C_2=(C_1V_1)/V_2` (and the RHS clearly has the units of concentration, why?)

So `C_2=(100.0*cancel(mL)xx0.150*mol*L^-1)/(150*cancel(mL))=0.100*mol*L^-1`

Here, I did not have to convert the `mL` volume to litres, because these clearly cancel out in the calculation to give an answer in `mol*L^-1` as required.........