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?

1 Answer
Jul 14, 2017

Answer:

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#