Molarity Problems Worksheet M = _n_ - n= # moles V - V must be in liters (change if necessary) - Use M or mol/L as unit for molarity 1. What is the molarity of a 0
Name _____ Period _____ MOLARITY WORKSHEET #1 For each of the following problems, use proper units and show ALL work:
1 Solutions Molarity Worksheet Name: _____KEY_____ 1. When we dissolve a cube of sugar in one cup of water, we create a homogeneous mixture
Molarity Worksheet 1 1. What does molarity mean? 2. What is the molarity of a solution that contains 4.53 moles of lithium nitrate in 2.85 liters of solution?
Homework Answers Molarity & Molality Worksheet Student Due Date 3/29/10 Concentration Problems: (Show your work for full credit!) 1) If you dissolve 20.0 g of NaOH in
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Molarity Practice Problems 1) ... Molarity Practice Problems – Answer Key 1) How many grams of potassium carbonate are needed to make 200 mL of a 2.5 M solution?
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Molarity Worksheet
W 331
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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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
Molarity Worksheet
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