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

1 Answer
anor277
Sep 12, 2017

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#.

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