In all cases below, be sure to explain your answers and show all work. When need

Question

In all cases below, be sure to explain your answers and show all work. When needed, you may assume the density of the solution is the same as the density of water: 1.00 g/mL You should have five questions on this post-lab assignment. A 2.85 g sample of groundwater was found to contain 1.63 micrograms of zinc ion. What is the concentration of zinc in mg'L and in ppm? Sodium fluoride (NaF) is one of the compounds used to fluoridate water. A community decides to supply its citizens w ith fluoridated w ater using a concentration of 1.0 ppm of NaF. How many grams of NaF must be added to the water in a 1.41 million gallon storage tank in order to reach this concentration? (946 mL = 1 qt. 4 qt = 1 gal) Death can result from ingestion of between 5 g and 10 g of fluoride. How many gallons of fluoridated water, with a fluoride concentration of 1.4 ppm, docs a person have to drink at one time to get 7.7 g of fluoride? (946 mL = 1 qt. 4 qt - 1 gal) What is the molarity of an NaF solution that is 152 ppm? A stock Br solution is prepared that is 6.52 ppm. 12.8 mL of water are added to 6.19 mL of this stock solution What is the Br concentration, in ppm, of the resulting solution?

Explanation / Answer

Question 1 )

So, 2.85 g of ground water

No of moles of water = 2.85/18 = 0.1583 moles

Mass of zinc = 1.63 micrograms = 1.63 *10^-6 g

Molar mass = 65 g/mol

No of moles of Zn = (1.63 *10^-6 )/65 = 2.5 * 10^-8 moles

So,

Concentration = 1.63*10^-3 mg / 2.85* 10^-3 L = 1.63 mg/L

OR

Concentration =  ((2.5 * 10^-8) /   0.1583)*10^6 = 0.15792 ppm

QUESTION 2 )

1 ppm of NaF

Volume = 1.41 million gallon

= 1.41 *4*946 * 10^6 mL = 5335.44 *10^6    mL

Density = 1g/mL

Mass of water =  5335.44 *10^6 g

Moles =  5335.44 *10^6 / 18 = 296.41 *10^6 moles

C = 1ppm,

1 = ( x/296.41 *10^6 )*10^-6

x = (296.41 *10^6 )/10^6 = 296.41 moles

Hmols of NaF required = 296.41 moles

Molecular mass of NaF = 42 g/mol

So, mass of NaF must be added = 296.41*42 grams = 12449.22 grams

QUESTION 3 )

C = (Moles of F-/Moles of water)*10^6 ppm = 1.4 ppm

Mass of F- required = 7.7 g

Molar mass of F- = 19 g/mole

So, moles of F- = 7.7/19 = 0.4052 moles

Since,

C = (Moles of F-/Moles of water)*10^6 ppm = 1.4 ppm

1.4 = (0.4052 / x) *10^6

x =  (0.4052 / 1.4) *10^6 = 0.2894 *10^6 moles

No of moles of water =  0.2894 *10^6 moles

molar mass of water = 18 g/mole

So, mass of water required = 18 *  0.2894 *10^6 grams = 5.2092 *10^6 grams

Density = 1g/mL

So,

Volume of water =  5.2092 *10^6 mL

=   5.2092 *10^6 / 946 qt

= 5.5 *10^3 qt

=  5.5 *10^3 / 4 gallon

= 1.375  *10^3 gallon

= 1375 gallon of water

So person has to drink  1375 gallon of water

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