Question

How to express 0.0620M aqueous solution of fluoride in mass percentage and parts per million

Answer 1

To express a 0.0620 M (mol/L) aqueous solution of fluoride in terms of mass percentage and parts per million (ppm), we need to follow a few steps.

### Step 1: Determine the Molar Mass of Fluoride

Assuming "fluoride" refers to the fluoride ion (F⁻), the molar mass of fluorine (F) is approximately 19.00 g/mol.

### Step 2: Convert Molarity to Mass Concentration

A 0.0620 M solution means there are 0.0620 moles of fluoride in 1 liter (1000 mL) of solution. To convert this to mass concentration:

\[
\text{mass of F} = \text{moles} \times \text{molar mass} = 0.0620 \, \text{mol} \times 19.00 \, \text{g/mol} = 1.178 \, \text{g}
\]

So, in 1 liter (1000 mL) of solution, there are approximately 1.178 grams of fluoride.

### Step 3: Calculate the Mass of the Solution

Assuming the density of the aqueous solution is approximately 1.00 g/mL (which is a reasonable approximation for dilute solutions):

\[
\text{mass of 1 liter of solution} = 1000 \, \text{mL} \times 1.00 \, \frac{\text{g}}{\text{mL}} = 1000 \, \text{g}
\]

### Step 4: Calculate Mass Percentage

The mass percentage can be calculated as:

\[
\text{Mass \%} = \left( \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \right) \times 100
\]
\[
\text{Mass \%} = \left( \frac{1.178 \, \text{g}}{1000 \, \text{g}} \right) \times 100 \approx 0.1178 \, \%
\]

### Step 5: Calculate Parts Per Million (PPM)

Parts per million (ppm) is calculated as:

\[
\text{ppm} = \left( \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \right) \times 10^6
\]
\[
\text{ppm} = \left( \frac{1.178 \, \text{g}}{1000 \, \text{g}} \right) \times 10^6 \approx 1178 \, \text{ppm}
\]

### Summary

- Mass percentage of fluoride in the solution: **0.1178%**
- Parts per million (ppm) of fluoride in the solution: **1178 ppm**