# How many moles of sodium hydroxide are required to prepare a 447.1*mL volume of a solution that is 1.20*mol*L^-1 in the solute?

Jul 14, 2017

We need a bit over $0.5 \cdot m o l$..............

#### Explanation:

$\text{Molarity"="Moles of solute"/"Volume of solution}$

And thus $\text{moles of solute"="volume of solution"xx"molarity}$

Is this clear? I am doing this problem dimensionally so that I know when to divide and when to multiply.........

And so we take the product..........(given that $1 \cdot m L \equiv {10}^{-} 3 L$)...

$447.1 \times {10}^{-} 3 \cdot \cancel{L} \times 1.20 \cdot m o l \cdot \cancel{{L}^{-} 1} = 0.537 \cdot m o l$.

We can even be more explicit than this and quote the mass of $N a O H$ required for this solution........

$\text{Mass of NaOH} = 0.537 \cdot m o l \times 40.0 \cdot g \cdot m o {l}^{-} 1 = 21.5 \cdot g$