Concentration Worksheet | Everett Community College, Slides of Chemistry

Find the mole fraction of the sodium chloride and of the water in the solution. 2). How many grams of magnesium cyanide are needed to make 275 mL of a 0.075. M ...

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Concentration Worksheet
W 328
Everett Community College
Student Support Services Program
1) 6.80 g of sodium chloride are added to 2750 mL of water. Find the mole fraction
of the sodium chloride and of the water in the solution.
2) How many grams of magnesium cyanide are needed to make 275 mL of a 0.075
M solution?
3) How many grams of magnesium cyanide would you need to add to 275 mL of
water to make a 0.075 molal solution?
4) Explain how to make one liter of a 1.25 molal sodium hydroxide solution.
5) What is the molarity of a solution made when 52 grams of potassium sulfate are
diluted to a volume of 4100 mL?
6) The density of ethylene glycol (antifreeze, HOCH2CH2OH) is 1.09 g/mL. How
many grams of ethylene glycol should be mixed with 375 mL of water to make a
7.50% (v/v) mixture?
7) Find the volume of a 0.75 M solution if it contains 39 grams of potassium
hydroxide.
8) How many grams of hydrochloric acid are present in 3.0 L of a 0.750 M solution?
9) The concentration of oxygen in water at the bottom of a lake is 0.48 g/L and the
pressure is 2.5 atm. If water from the bottom is moved by a current upwards to
a depth where the pressure is 1.3 atm, what is the concentration of the oxygen
in the water at this depth?
10) What is the molarity of a solution in which 0.850 grams of ammonium nitrate are
dissolved in 345 mL of solution?
11) Explain how you would make 675 mL of a 0.400 M barium iodide solution.
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Concentration Worksheet

W 328

Everett Community College Student Support Services Program

1) 6.80 g of sodium chloride are added to 2750 mL of water. Find the mole fraction

of the sodium chloride and of the water in the solution.

2) How many grams of magnesium cyanide are needed to make 275 mL of a 0.

M solution?

3) How many grams of magnesium cyanide would you need to add to 275 mL of

water to make a 0.075 molal solution?

4) Explain how to make one liter of a 1.25 molal sodium hydroxide solution.

5) What is the molarity of a solution made when 52 grams of potassium sulfate are

diluted to a volume of 4100 mL?

6) The density of ethylene glycol (antifreeze, HOCH 2 CH 2 OH) is 1.09 g/mL. How

many grams of ethylene glycol should be mixed with 375 mL of water to make a

7.50% (v/v) mixture?

7) Find the volume of a 0.75 M solution if it contains 39 grams of potassium

hydroxide.

8) How many grams of hydrochloric acid are present in 3.0 L of a 0.750 M solution?

9) The concentration of oxygen in water at the bottom of a lake is 0.48 g/L and the

pressure is 2.5 atm. If water from the bottom is moved by a current upwards to

a depth where the pressure is 1.3 atm, what is the concentration of the oxygen

in the water at this depth?

10) What is the molarity of a solution in which 0.850 grams of ammonium nitrate are

dissolved in 345 mL of solution?

11) Explain how you would make 675 mL of a 0.400 M barium iodide solution.

1. 6.80 g NaCl x 1 mole NaCl = 0.116 mole NaCl 58.45 g NaCl 2750 mL H 2 O x 1 g H 2 O x 1 mole H 2 O = 152.8 mol H 2 O 1 mL H 2 O 18 g H 2 O mole fraction NaCl = 0.116 mole NaCl = 7.59 x 10- 152.9 mole soln mole fraction H 2 O = 152.8 mole H 2 O = 0. 152.9 mole soln 2. 275 mL H 2 O x 1 g H 2 O = 0.275 L H 2 O 1000 mL H 2 O 275 L Mg(CN) 2 x 0.075 mole Mg(CN) 2 x 76.3 g Mg(CN) 2 = 1.6 g Mg(CN) 2 1 L Mg(CN) 2 1 mole Mg(CN) 2 3. 275 mL H 2 O x 1 g H 2 O x 1 kg H 2 O = 0.275 kg H 2 O 1 mL H 2 O 1 g H 2 O 0.275 kg H 2 O x 0.075 mol Mg(CN) 2 x 76.3 g Mg(CN) 2 = 1.6 g Mg(CN) 2 1 kg H 2 O 1 mole Mg(CN) 2 4. 1.25 molal NaOH = 1.25 mole NaOH 1 kg H 2 O 1.25 mole NaOH x 40 g NaOH = 50.0 g NaOH 1 mole NaOH Measure 50.0 g NaOH and add water to 1 L volume. 5. 52 g K 2 SO 4 x 1 mole K 2 SO 4 = 0.299 mole K 2 SO 4 174 g K 2 SO 4 0.299 mole K 2 SO 4 = 0.073 M 4.100 L K 2 SO 4 6. 375 mL x 0.0750 = 28.125 mL ethylene glycol 28.125 mL ethylene glycol x 1.09 g ethylene glycol/1ml = 30.7 g ethylene glycol 7. 39 g KOH x 1 mole KOH x 1 L KOH = 0.93 L = 930 mL 56 g KOH 0.75 mol KOH 8. 3.0 L soln x 0.750 moles HCl x 36.45 g HCl = 82 g HCl 1 L soln 1 mole HCl 9. 1.3 atm = C  C = (1.3 atm)(0.48 g/L) = 0.25 g/L 2.5 atm 0.48 g/L (2.5 atm) 10. 0.850 g NH 4 NO 3 x 1 mole NH 4 NO 3 = 0.0106 mole NH 4 NO 3 80 g NH 4 NO 3 0.0106 mole NH 4 NO 3 = 0.0307 M 0.345 L NH 4 NO 3 11. 0.675 L BaI 2 x 0.400 moles BaI 2 = 0.270 moles BaI 2

1 L BaI 2

0.270 moles BaI 2 x 391.1 g BaI 2 = 106 g BaI 2 1 mole BaI 2 Measure 106 g BaI 2 into a beaker and add water to 675 mL volume.