How many milligrams of sodium fluoride are needed to prepare 100 ml of a sodium fluoride stock solution such that a solution containing 2 ppm?

The unit ppm (parts per million) should be numerically equal to the unit of mg/L (milligram per liter). First, we determine the volume of the stock solution in L.

V = (100 mL) (1 L/1000 mL)

V = 0.1 L

Then, the equation that would allow us to determine the value of mass is,

C = m/V

where C is concentration, m is mass, and V is the volume.

Substituting the known values,

2 = m / 0.1

The value of m from the equation is 0.2.

Answer: 0.2 mg