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

Sep 12, 2017

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

#### Explanation:

We require $\frac{6.22 \cdot g}{40.00 \cdot g \cdot m o {l}^{-} 1} = 0.0156 \cdot m o l$ WITH RESPECT TO $N a O H$.

Now, by definition, $\text{Concentration"="Moles"/"Volume}$......

And thus $\text{Volume"="Moles"/"Concentration} = \frac{0.0156 \cdot m o l}{0.315 \cdot m o l \cdot {L}^{-} 1}$

$= 0.494 \cdot L$.