University:
Eastern Kentucky University
Course:
CHE 111 | General Chemistry
Academic year:

2020
Views:

90

Pages:

2
Author:

Carina Becker

Molarity Worksheet 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 (NH4)2CO3 x 1 mole (NH4)2CO3 = 9.92 mole (NH4)2CO3 96 g (NH4)2CO3 9.92 mole (NH4)2CO3 = 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(NO3)4 x 1 mole Pb(NO3)4 = 0.0216 moles Pb(NO3)4 455.2 g Pb(NO3)4 0.0216 moles Pb(NO3)4 = 0.0465 M 0.0465 L soln