Molarity Worksheet | Everett Community College, Study notes of Chemistry

What is the molarity of the following solutions given that: 1). 1.0 moles of potassium fluoride is dissolved to make 0.10 L of solution. 1.0 mole KF = 10. M.

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Molarity Worksheet
W 331
Everett Community College
Student Support Services Program
What is the molarity of the following solutions given that:
1) 1.0 moles of potassium fluoride is dissolved to make 0.10 L of solution.
2) 1.0 grams of potassium fluoride is dissolved to make 0.10 L of solution.
3) 1.0 grams of potassium fluoride is dissolved to make 0.10 mL of solution.
4) 952 grams of ammonium carbonate are dissolved to make 1750 mL of solution.
5) 9.82 grams of lead (IV) nitrate are dissolved to make 465 mL of solution.
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Molarity Worksheet

W 331

Everett Community College Student Support Services Program What is the molarity of the following solutions given that:

  1. 1.0 moles of potassium fluoride is dissolved to make 0.10 L of solution.
  2. 1.0 grams of potassium fluoride is dissolved to make 0.10 L of solution.
  3. 1.0 grams of potassium fluoride is dissolved to make 0.10 mL of solution.
  4. 952 grams of ammonium carbonate are dissolved to make 1750 mL of solution.
  5. 9.82 grams of lead (IV) nitrate are dissolved to make 465 mL of solution.

Solutions What is the molarity of the following solutions given that:

  1. 1.0 moles of potassium fluoride is dissolved to make 0.10 L of solution. 1.0 mole KF = 10. M 0.10 L soln
  2. 1.0 grams of potassium fluoride is dissolved to make 0.10 L of solution. 1.0 g KF x 1 mole KF = 0.0172 mol KF 58 g KF 0.0172 mol KF = 0.17 M 0.10 L soln
  3. 1.0 grams of potassium fluoride is dissolved to make 0.10 mL of solution. 1.0 g KF x 1 mole KF = 0.0172 mol KF 58 g KF 0.0172 mol KF = 170 M 1 x 10-4^ L soln
  4. 952 grams of ammonium carbonate are dissolved to make 1750 mL of solution. 952 g (NH 4 ) 2 CO 3 x 1 mole (NH 4 ) 2 CO 3 = 9.92 mole (NH 4 ) 2 CO 3 96 g (NH 4 ) 2 CO 3 9.92 mole (NH 4 ) 2 CO 3 = 5.67 M 1.75 L soln
  5. 9.82 grams of lead (IV) nitrate are dissolved to make 465 mL of solution. 9.82 g Pb(NO 3 ) 4 x 1 mole Pb(NO 3 ) 4 = 0.0216 moles Pb(NO 3 ) 4 455.2 g Pb(NO 3 ) 4 0.0216 moles Pb(NO 3 ) 4 = 0.0465 M 0.0465 L soln