What volume of $0.315molL^-1$ $NaOH$ is required to deliver $6.22*g$ of $NaOH$ ?

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Answer 1 · 2017-09-12T08:14:52

Approx. half a litre of the given solution is required.........

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 $"Volume"="Moles"/"Concentration"=(0.0156*mol)/(0.315*mol*L^-1)$

$=0.494*L$ .

What volume of 0.315*mol*L^-10.315*mol*L^-1 NaOHNaOH is required to deliver 6.22*g6.22*g of NaOHNaOH?