Question

How many moles of sodium hydroxide are required to prepare a `447.1*mL` volume of a solution that is `1.20*mol*L^-1` in the solute?

Answer 1

We need a bit over `0.5*mol`..............

Explanation:

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

And thus `"moles of solute"="volume of solution"xx"molarity"`

Is this clear? I am doing this problem dimensionally so that I know when to divide and when to multiply.........

And so we take the product..........(given that `1*mL-=10^-3L`)...

`447.1xx10^-3*cancelLxx1.20*mol*cancel(L^-1)=0.537*mol`.

We can even be more explicit than this and quote the mass of `NaOH` required for this solution........

`"Mass of NaOH"=0.537*molxx40.0*g*mol^-1=21.5*g`