Question

How many grams of NaF are needed to make 6.3 liters of a 3.6
molar solution?

Answer 1

9.7 × 10² g NaF.

Step-by-step explanation:

Molarity is defined by the amount of solute (mols) over the volume of solution (liters):

`\displaystyle \text{Molarity} = \frac{\text{mols solute}}{\text{L solution}}`

We want to create 6.3 liters of a 3.6 molar solution of NaF.

Solve for the amount of NaF necessary:

`\displaystyle \begin{aligned} (3.6\text{ M}) & = \frac{\text{ mol NaF}}{6.3\text{ L}} \\ \\ \text{mol NaF} & = 23 \text{ mol}\end{aligned}`

Therefore, about 23 moles of NaF is required.

Convert from moles to grams. The molecular weight of NaF is 41.99 g/mol:

`\displaystyle 23\text{ mol NaF} \cdot \frac{41.99\text{ g NaF}}{1\text{ mol NaF}} = 9.7* 10^2\text{ g NaF}`

Therefore, about 970 grams of NaF is needed to create the solution.