# What volume of "lithium sulfite solution" of 2.50*mol*L^-1 concentration would contain a mass of 422.55*g?

Apr 17, 2017

$\text{Moles of lithium sulfite req'd...........} = \frac{422.55 \cdot g}{93.94 \cdot g \cdot m o {l}^{-} 1} =$

#### Explanation:

$= \frac{422.55 \cdot g}{93.94 \cdot g \cdot m o {l}^{-} 1} = 4.50 \cdot m o l$.

And since $\text{concentration"="moles"/"volume}$, we need the quotient..........

$\text{volume"="moles"/"concentration} = \frac{4.50 \cdot m o l}{2.50 \cdot m o l \cdot {L}^{-} 1} = 1.80 \cdot \frac{1}{L} ^ - 1$

$1.80 \cdot \frac{1}{\frac{1}{L}} = 1.80 \cdot L$

In practice, water is quite an involatile material. You would not evaporate this quantity of water.