# How do you convert 0.45 g of sodium hydroxide, NaOH to moles?

We take the quotient...${n}_{\text{NaOH"="mass of sodium hydroxide"/"molar mass of sodium hydroxide}}$
${n}_{\text{NaOH}} = \frac{0.450 \cdot g}{40.00 \cdot g \cdot m o {l}^{-} 1} = 0.0113 \cdot m o l$
Note the dimensional consistency of the answer. The $g$ cancel...and we get an answer in $\frac{\cancel{g}}{\cancel{g} \cdot m o {l}^{-} 1} = \frac{1}{m o {l}^{-} 1} = \frac{1}{\frac{1}{m o l}} = m o l$...as required...