# With respect to NaOH(s), what is the concentration of 15.5*g mass of sodium hydroxide dissolved in a 650*mL volume of water?

May 4, 2017

$\text{Molarity"="Moles of solute"/"Volume of solution} \cong 0.6 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

Given the formula:

"Molarity"=((15.50*cancelg)/(40.00*cancelg*mol^-1))/(650.0*cancel(mL)xx10^-3*L*cancel(mL^-1))=??*mol*L^-1.

Note that this is dimensionally consistent, i.e. $\frac{1}{m o {l}^{-} 1} = \frac{1}{\frac{1}{m o l}}$ $= m o l$ as required...........

Since we wanted an answer in $m o l \cdot {L}^{-} 1$ expressing $\text{concentration}$, and we got one, this is a good check that we performed the analysis properly. It is all too easy to divide instead of multiply or vice versa, and botch the problem entirely. We have all done it.