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Ch 13 Solutions Sec 2 pg 467 #1-11

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1. Why did chemists develop the concept of molarity? To easily connect the amount in moles of a substance with its concentration in a solution

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2. In what units is molarity expressed Moles per liter Mol/L

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3. Describe in your own words how to prepare mL of a 0.85 M solution of sodium chloride. 23Na + 36Cl= 59NaCl.85 mol NaCl x 59 g/mol = 50 g/L or 5.0g NaCl for 1/10 L (100ml) Weigh 5.0 g of salt, dissolve in water. Pour into 100 ml volumetric flask, swirl and fill to volumetric line

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4. If you dissolve 2.00 mol KI in 1.00 L of water, will you get 2.00 M solution? Explain. No The total volume may be different by the time the solute is dissolved.

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5. A sample of g of water is found to contain 175 mg Cd. What is this concentration in parts per million. Change mg to g 175 mg x 1 g/ 1000 mg =.175 g Cd Find g Cd/g of water.175 g/ g water x 1000,000 parts /1 million = or 438 parts per million or ppm

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6. If 1.63 x g of helium dissolves in g of water, what is the concentration in parts per million? How much in 1 gram of water? 1.63 x g He / 100g H2O x 1000,000 parts/million 1.63 parts per million or ppm He

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A standard solution of NaOH is M. What mass of NaOH is present in mL of the solution? Na = 23, O = 16, H = 1 NaOH = g in 1 M 100 mL is 1/10 L so 40/10 = 4 g in mL

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8. A 32 g sample of LiCl is dissolved in water to to form 655 mL of solution. What is the molarity of the solution? Convert mass to moles. Li = 6.9,Cl = g x mol/43g LiCl =.74 mol LiCl Convert mL to L 655 ml x 1 L/ 1000mL =.655 L Find Moles per Liter.74 mol /.655 L = or 1.1 M LiCl

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9. Most household bleach contains sodium hypochlorite, NaOCl. A 2.84 L bottle contains 177 g NaOCl. What is the molarity of the solution. Find moles of NaOCl Na=23,O=16,Cl= g NaOCl x mol/74.5g NaOCl = 2.38 Mol NaOCl Find mol/L 2.38 mol/2.84 L =.8365 or.837 mol/L or M

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10. What mass of AgNO3 is needed to prepare mL of a M solution. Ag=108,N=14,O3= 16(3)= 48 AgNO3 = 170 g M = g/L = 170g /mol = for 1 Mol solution 170/4 for.250 L = 42.5 g for.250 L AgNO g for 1 M, 42.5 x.125 M 5.25 or 5.3 g AgNO3 for.125 M solution

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11. Calcium phosphate used in fertilizers can be made in the reaction described by the following equation: 2H 3 PO 4 (aq) + 3Ca(OH) 2 (aq) --> Ca 3 (PO 4 ) 2 (s) + 6H 2 O(aq) Ca=40(3), P=31(2), O=16(8)=128, = 310g total H= 1(2), O=16, total 18g total What mass in grams of each product would be formed if 7.5 L of 5.00 M phosphoric acid reacted with an excess of calcium hydroxide? H3PO4 = 1(3) (4) or 64 = 98g for 1 M 98x5 x 7.5 = 3,562.5 or 3600 g H3PO4 3600g 2H 3 PO 4 x 1 mol/98g x 1 mol Ca 3 (PO 4 ) 2 /2mol 2H 3 PO 4 x 310 g Ca 3 (PO 4 ) 2 /1 mol Ca 3 (PO 4 ) 2 = 5,693.8 or 5.7 x 10 3 Ca 3 (PO 4 ) 2 18g H2O

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