# What mass of lithium hydroxide is need to prepare a 250*mL volume of "LiOH" at 0.33*mol*L^-1 concentration?

Oct 24, 2016

We require an approximate mass of $2.0 \cdot g$ of $L i O H$.
$\text{Concentration}$ $=$ $\text{Moles of solute"/"Litres of solution}$. And thus the typical units of concentration are $m o l \cdot {L}^{-} 1$.
We want to make a $250 \cdot m L$ volume of $0.33 \cdot m o l \cdot {L}^{-} 1$ concentration. And thus we need $0.250 \cdot L \times 0.33 \cdot m o l \cdot {L}^{-} 1$ $=$ $0.0825 \cdot m o l$ of $L i O H$ solute.
And this molar quantity has a mass of $0.0825 \cdot m o l \times 23.95 \cdot g \cdot m o {l}^{-} 1$ $=$ $1.98 \cdot g$