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

Question

1563 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-14T20:38:31

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

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

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........

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

How many moles of sodium hydroxide are required to prepare a 447.1*mL447.1*mL volume of a solution that is 1.20*mol*L^-11.20*mol*L^-1 in the solute?