Time to CONCENTRATE!
1.
A sample of 500.0 g of drinking water is found to contain 44 mg of arsenic. What is this concentration in
parts per million (ppm)?
44 ÷ 1000 = .044 g
(.044 ÷ 500.0) x 1,000,000 = 88 ppm
2.
A 750.0 g soft-water sample contains 151 mg Na. What is this concentration in ppm?
3.
A 600.0 g sample of paint has 22.2 mg of “Yellow #5.” What is this concentration in ppm?
4.
A solution is being made with 65.5 g of NaCl dissolved in 2,250 mL of water.
151 ÷ 1000 = .151 g
(.151 ÷ 750.0) x 1,000,000 = 201 ppm
22.2 ÷ 1000 = .0222 g
(.0222 ÷ 600.0) x 1,000,000 = 37.0 ppm
a)
How many liters is 2,250 mL?
b)
How many moles is 65.5 g of NaCl?
2,250 mL ÷ 1000 = 2.25 L
mass ÷ molar mass (PT) = moles
65.5 ÷ 58.5 = 1.12 mol
58.5 = 23.0 (Na) + 35.5 (Cl)
c)
5.
What is the Molarity of the solution?
1.12 mol ÷ 2.25 L = 0.498 M
A solution is being made with 4471 g of Al2S3 dissolved in 5,750 mL of water.
a)
How many liters is 5,750 mL?
b)
How many moles is 4471 g of Al2S3?
5,750 mL ÷ 1000 = 5.75 L
mass ÷ molar mass (PT) = moles
4471 ÷ 150.3 = 29.75 mol
150.3 = 54.0 (2 Al, 27.0) + 96.3 (3 S, 32.1)
c)
What is the Molarity of the solution?
29.75 mol ÷ 5.75 L = 5.17 M
6.
How many moles of sugar must be dissolved in 1.5 Liters of water to make a 4.0 M solution?
4.0 M = moles
1.5 L
7.
So, 4.0 x 1.5 = 6.0 moles
How many grams of CaBr2 must be dissolved in 2.25 Liters of water to make a 0.335 M solution?
0.335 M = moles
2.25 L
So, 0.335 x 2.25 = 0.75375 moles
Then, moles x molar mass (PT) = grams
0.75375 x 199.9 = 151 g
(199.9 = Ca (40.1) + Br2 (79.9 + 79.9) = 40.1 + 159.8)
8.
If 315 g of C6H12O6 is dissolved in enough water to make 750. mL of solution, what is the molarity of the
sugar solution?
mass ÷ molar mass (PT) = moles
315 ÷ 180.0 = 1.75 moles (180.0 = 6C + 12H + 6O = 6(12.0) + 12(1.0) + 6(16.0) = 72+12+96)
Then, mL ÷ 1000 = L, so 750. ÷ 1000 = .750 L
Finally, 1.75 mol ÷ .750 L = 2.33 M
9.
What mass of Na2CO3 is present in 425 mL of a 0.75 M solution of sodium carbonate?
0.75 M = moles
.425 L
425 mL ÷ 1000 = .425 L
So, 0.75 x .425 = 0.31875 moles
Then, moles x molar mass (PT) = grams
0.31875 x 106.0 = 34 g
(106.0 = 2Na + 1C + 3O = 2(23.0) + 1(12.0) + 3(16.0) = 46+12+48)
10.
If 250.0 L of a 1.5 M solution of NaCl evaporates, what mass of sodium chloride will remain behind?
1.5 M = moles
250. L
So, 1.5 x 250. = 375 moles
Then, moles x molar mass (PT) = grams
375 x 58.5 = 22,000 g
58.5 = 23.0 (Na) + 35.5 (Cl)
11.
Mr. Senn needs to make a phosphoric acid solution for a lab. He has 10 M H3PO4 in stock, but he needs
1200. mL of 4.5 M H3PO4 for the lab. HOW DOES HE DO THIS? *Don’t forget the sentence!!
M1V1 = M2V2
(10)(V1) = (4.5)(1200)
10x = 5400
x = 540 mL
Take 540 mL of 10 M H3PO4 and dilute it with 660 mL of water.