advertisement

Ch 13 Solutions Sec 2 pg 467 #1-11 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 2. In what units is molarity expressed • Moles per liter • Mol/L 3. Describe in your own words how to prepare 100.0 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 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. 5. A sample of 400.0 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/ 400.0 g water x 1000,000 parts /1 million = • 437.5 or 438 parts per million or ppm 6. If 1.63 x 10-4 g of helium dissolves in 100.0 g of water, what is the concentration in parts per million? • How much in 1 gram of water? • 1.63 x 10-4 g He / 100g H2O x 1000,000 parts/million • 1.63 parts per million or ppm He A standard solution of NaOH is 1.000 M. What mass of NaOH is present in 100.0 mL of the solution? • • • • Na = 23, O = 16, H = 1 NaOH = 40 40 g in 1 M 100 mL is 1/10 L so 40/10 = 4 g in 100.0 mL 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 = 36 • 32 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 = 1.129 or 1.1 M LiCl 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=35.5 • 177g 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 10. What mass of AgNO3 is needed to prepare 250.0 mL of a 0.125 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 AgNO3 42.5 g for 1 M, 42.5 x .125 M 5.25 or 5.3 g AgNO3 for .125 M solution in the reaction described by the following equation: 2H3PO4(aq) + 3Ca(OH)2(aq) --> Ca3(PO4)2(s) + 6H2O(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) + 31 + 16(4) or 64 = 98g for 1 M • 98x5 x 7.5 = 3,562.5 or 3600 g H3PO4 • 3600g 2H3PO4 x 1 mol/98g x 1 mol Ca3(PO4)2 /2mol 2H3PO4 • x310 g Ca3(PO4)2 /1 mol Ca3(PO4)2 = 5,693.8 or 5.7 x 103 Ca3(PO4)2 • 18g H2O