# What is the volume of a 3.00 M solution made with 122 g of LiF?

## Question:

What is the volume of a {eq}3.00\ M {/eq} solution made with {eq}122\ g {/eq} of {eq}\rm LiF {/eq}?

## Molarity:

The molarity is an expression of the concentration of the solution in terms of the number of moles, {eq}\displaystyle n {/eq}, of a reagent per unit volume, {eq}\displaystyle V {/eq}, of the solution. We can, therefore, acquire the molarity by dividing the number of moles by the total volume of the solution, or {eq}\displaystyle M = \frac{n}{V} {/eq}.

Determine the volume, {eq}\displaystyle V {/eq}, of the given solution by first finding the number of moles there is in the given mass, {eq}\displaystyle m = 122\ g\ LiF {/eq}, by its molar mass, {eq}\displaystyle 25.939 \rm{g/mol} {/eq}, and then dividing it by the molarity of the solution, {eq}\displaystyle M = 3.00\ M {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle V &= \frac{n}{M}\\ &= \frac{\frac{122\ g\ LiF}{25.939\ \rm{g/mol}}}{3.00\ M}\\ &\approx 1.57\ L \end{align} {/eq} 