# What volume (in mL) of a 3.50M LiF stock solution would you use to make 1.5L of 0.500M LiF solution?

Aug 20, 2016

A volume a bit over $214 \cdot m L$ is required.

#### Explanation:

$\text{Moles of lithium fluoride}$ $=$ $1.5 \cdot L \times 0.500 \cdot m o l \cdot {L}^{-} 1 = 0.75 \cdot m o l$.

And thus we need a volume of the stock solution that contains $0.75 \cdot m o l$ lithium fluoride.

$\frac{0.75 \cdot \cancel{m o l}}{3.50 \cdot \cancel{m o l} \cdot \cancel{{L}^{-} 1}} \times {10}^{3} \cdot m L \cdot \cancel{{L}^{-} 1} = 214.3 \cdot m L$

This solution would be slightly basic, i.e. $p H > 7$. Why?