Molarity Calculation practice. #1 M = mol L What is the concentration of a solution with 0.25 mol of solute in 0.75 L of solution? M = 0.25 mol = 0.33.

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Molarity Calculation practice

#1 M = mol L What is the concentration of a solution with 0.25 mol of solute in 0.75 L of solution? M = 0.25 mol = 0.33 M 0.75 L

#2 M = mol L What is the concentration of a solution with 0.35 mol of solute in 1.75 L of solution? M = 0.35 mol = 0.20 M 1.75 L

#3 M = mol L How many moles of solute are in 1.5 L of a 0.01 M solution? 0.01M = x mol 1.5 L X = mol solute

#4 M = mol L How many moles of calcium chloride are in 0.50 L of a 0.20 M solution? 0.20M = x mol 0.50 L X = 0.10 mol CaCl 2

#5 M = mol L How many grams of calcium chloride are in 0.50 L of a 0.20 M solution? 0.20M = x mol 0.50 L X = 0.10 mol CaCl mol CaCl 2 x g = g = 11 g CaCl 2 1 mol

#5 continued Explain (in ~2 sentences) how to make the solution in the previous example. 1.Weigh out 11 g CaCl 2 and pour it into a 500 mL volumetric flask. 2.Add enough distilled water to dissolve the solute—mix well. 3.Add water to fill the volumetric flask to the measurement line and mix completely by inverting.

#6 M = mol L How many grams of potassium nitrate are in 250 mL of a 0.15 M solution? 0.15M = x mol 0.25 L X = mol KNO mol KNO 3 x g = g = 3.8 g KNO 3 1 mol 250 mol = x L. 1000mL x = L

#6 continued Explain (in ~2 sentences) how to make the solution in the previous example. 1.Weigh out 3.8 g KNO 3 and pour it into a 250 mL volumetric flask. 2.Add enough distilled water to dissolve the solute—mix well. 3.Add water to fill the volumetric flask to the measurement line and mix completely by inverting.

#7 M = mol L If 10.0 grams of strontium nitrate are dissolved in 300. mL of solution, what is the concentration? 10.0 g Sr(NO 3 ) 2 x 1 mol = mol Sr(NO 3 ) g M= mol= M Sr(NO 3 ) L 300 mL = x L mL X = L

#8 M = mol L mol CaCl 2 x 2 mol AgCl x g AgCl = g 1 mol CaCl 2 1 mol AgCl AgCl CaCl 2 (aq) + 2 AgNO 3 (aq)  Ca(NO 3 ) 2 (aq) + 2 AgCl (s) 500. mL = x L. 1000mL x = L 0.20M = x mol L x = mol CaCl 2 How many grams of silver chloride will be produced if 500. mL of a M solution of CaCl 2 is mixed with 500. mL of a very concentrated solution of AgNO 3 ?

#8 M = mol L mol CaCl 2 x 2 mol AgCl x g AgCl = 28.7 g 1 mol CaCl 2 1 mol AgCl AgCl CaCl 2 (aq) + 2 AgNO 3 (aq)  Ca(NO 3 ) 2 (aq) + 2 AgCl (s) 500. mL = x L. 1000mL x = L 0.20M = x mol L x = mol CaCl 2 How many grams of silver chloride will be produced if 500. mL of a M solution of CaCl 2 is mixed with 500. mL of a very concentrated solution of AgNO 3 ?

#9 M = mol L 0.10 mol CaCl 2 x 2 mol AgCl x g AgCl = g 1 mol CaCl 2 1 mol AgCl AgCl CaCl 2 (aq) + 2 AgNO 3 (aq)  Ca(NO 3 ) 2 (aq) + 2 AgCl (s) 50. mL = x L. 1000mL x = L 2.0M CaCl 2 = x mol L x = 0.10 mol CaCl 2 How many grams of silver chloride will be produced if 50. mL of a 2.0 M solution of CaCl 2 is mixed with 50. mL of a 3.0 M solution of AgNO 3 ? 3.0M AgNO 3 = x mol L x = 0.15 mol AgNO mol AgNO 3 x 2 mol AgCl x g AgCl = g 2 mol AgNO 3 1 mol AgCl AgCl

#9 M = mol L 0.10 mol CaCl 2 x 2 mol AgCl x g AgCl = 28.7 g 1 mol CaCl 2 1 mol AgCl AgCl CaCl 2 (aq) + 2 AgNO 3 (aq)  Ca(NO 3 ) 2 (aq) + 2 AgCl (s) 50.0 mL = x L. 1000mL x = L 2.0M CaCl 2 = x mol L x = mol CaCl 2 How many grams of silver chloride will be produced if 50.0 mL of a 2.0 M solution of CaCl 2 is mixed with 50.0 mL of a 3.0 M solution of AgNO 3 ? 3.0M AgNO 3 = x mol L x = mol AgNO mol AgNO 3 x 2 mol AgCl x g AgCl = 21.5 g 2 mol AgNO 3 1 mol AgCl AgCl

#9 M = mol L 0.10 mol CaCl 2 x 2 mol AgCl x g AgCl = 28.7 g 1 mol CaCl 2 1 mol AgCl AgCl CaCl 2 (aq) + 2 AgNO 3 (aq)  Ca(NO 3 ) 2 (aq) + 2 AgCl (s) 50.0 mL = x L. 1000mL x = L 2.0M CaCl 2 = x mol L x = mol CaCl 2 How many grams of silver chloride will be produced if 50.0 mL of a 2.0 M solution of CaCl 2 is mixed with 50.0 mL of a 3.0 M solution of AgNO 3 ? 3.0M AgNO 3 = x mol L x = mol AgNO mol AgNO 3 x 2 mol AgCl x g AgCl = 21.5 g 2 mol AgNO 3 1 mol AgCl AgCl