What is the concentration of a 40*g mass of sodium hydroxide dissolved in 250*mL of water?

Jul 30, 2017

Approx. $4 \cdot m o l \cdot {L}^{-} 1$.......or $4.0 \cdot \text{normal}$

Explanation:

$\left[N a O H\right] = \frac{\frac{40 \cdot g}{40 \cdot g \cdot m o {l}^{-} 1}}{250 \times {10}^{3} \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1} = 4.0 \cdot m o l \cdot {L}^{-} 1$

This follows the normal means of calculation, $\text{concentration"="moles of solute"/"volume of solution}$, and thus get an answer in $m o l \cdot {L}^{-} 1$.

We assume no volume change on dissolution.

And under these circumstances, where the solute causes no further water hydrolysis, $\text{molarity "-=" normality}$...........