# A 16*g mass of sodium hydroxide is dissolved in enough water to give 2.0*L of solution. What is the concentration of the solution in mol*L^-1?

Apr 2, 2017

$\text{Molarity}$ $=$ $\text{Moles of solute"/"Volume of solution} =$
$\text{Molarity}$ $=$ $\frac{\frac{16.0 \cdot \cancel{g}}{40.0 \cdot \cancel{g} \cdot m o {l}^{-} 1}}{2.0 \cdot L} = 0.20 \cdot m o l \cdot {L}^{-} 1$.
Note how the units cancel to give the required units of concentration, i.e. $m o l \cdot {L}^{-} 1$:
$\frac{\frac{\cancel{g}}{\cdot \cancel{g} \cdot m o {l}^{-} 1}}{L} = m o l \cdot {L}^{-} 1$, i.e. $\frac{1}{x} ^ - 1$ $=$ $\frac{1}{\frac{1}{x}}$ $=$ $x$ as required...........