How many grams of NaOH are needed to prepare 250 mL of 0.205 M NaOH?

Feb 28, 2017

A bit over $2 \cdot g$...............

Explanation:

$\text{Concentration}$ $=$ $\text{Moles of solute (moles)"/"Volume of solution (L)}$

Thus $\text{Concentration"xx"volume"="moles of solute}$

$0.205 \cdot m o l \cdot \cancel{{L}^{-} 1} \times 0.250 \cdot \cancel{L} = \text{Moles of solute}$

$= 0.0513 \cdot m o l$, an answer in moles as required.

And thus $\text{mass of NaOH}$ $=$ $\text{moles "xx" molar mass}$

$= 0.0513 \cdot \cancel{m o l} \times 40.0 \cdot g \cdot \cancel{m o {l}^{-} 1}$

$= 2.05 \cdot g$