* You may assume solution densities are the same as the density of water: 1.00 g

Question

* You may assume solution densities are the same as the density of water: 1.00 g/mL.

Question:1
Perform the calculations to complete the table in the lab (i.e., calculate the concentration of each solution in ppm).



Question:2
How many micrograms of fluoride are present in 3.8 g of a solution that has a fluoride concentration of 8.3 ppm?



Question:3
Calculate the concentration, in ppm, of a solution prepared by dissolving 0.00401 grams of sodium fluoride in 1.88 L of pure water.



Question:4
Calculate the molar concentration of fluoride in a solution that is 24.4 ppm in fluoride.

Explanation / Answer

1)

recall that ppm = microgram of solute / kg of solution

2)

micrograms of F in

m = 3.8 g of solution with C = 8.3 ppm

ppm = microgram / kg solution

microgram of F = ppm * kg solution = (8.3)(3.8*10^-3) = 0.03154 micrograms of F in such 3.8 g of solution

3)

m = 0.00401 g of NaF = 0.00401*10^6 micrograms

V = 1.88 L of pure water

ppm = microgram of solute / Volume of solution in liter

ppm = (0.00401*10^6) / (1.88 L) = 2132.978 ppm

4)

molar concnetration of F- in solutino of 24.4 ppm in F

ppm = micrograms F / liter of solution

find micrograms

ppm*L = micrograms

L = 1 (molar concentration states M = 1 mol per liter)

24.4*1 = micrograms

24.4 micrograms of F- ions are expected

since

MW of F = 18.99 g/mol

thn

mol = mass/MW

mol = (24.4*10^-6) / (18.99) = 1.28*10^-6 mol (in 1 liter)

M = 1.28*10^-6 mol per liter

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