1. Hydrogen bonding… (a) occurs only between water molecules (b) is stronger than covalent bonding (c) can occur between NH 3 and H 2 O (d) results from strong attractive forces in ionic compounds

2. Which of the following is not a general property of solutions? (a) a homogeneous mixture of two or more substances (b) variable composition (c) dissolved solute breaks down to individual molecules (d) the same chemical composition, the same chemical properties, and the same physical properties in every part

3. Which procedure is most likely to increase the solubility of most solids in liquids? (a) stirring (b) pulverizing the solid (c) heating the solution (d) increasing the pressure

4. The addition of a crystal of NaClO 3 to a solution of NaClO 3 causes additional crystals to precipitate. The original solution was (a) unsaturated (b) dilute (c) saturated (d) supersaturated

5. If NaCl is soluble in water to the extent of 36.0 g NaCl / 100 g H 2 O at 20 o C, then a solution at 20 o C containing 45.0 g NaCl / 150 g H 2 O would be… 45.0 g NaCl = X g NaCl 150 g H 2 O100 g H 2 O X = 30.0 g NaCl is unsaturated

6. If 5.00 g NaCl is dissolved in 25.0 g H 2 O, the percent NaCl by mass is 5.00 g NaCl=5.00 5.00 g NaCl + 25.0 g H 2 O30.0 = 16.7 %

7. How many grams of 9.0% AgNO 3 solution will contain 5.3 g AgNO 3 ? 9.0 = 5.3 100 X X = 59

8. What mass of BaCl 2 will be required to prepare 200. mL of 0.150 M solution? 0.150 mol x 0.200 L = 0.0300 mol L x 208.3g/mol = 6.25 g BaCl 2

9. How many grams of a solution that is 12.5 % by mass AgNO 3 would contain 0.400 mol of AgNO 3 ? 0.400 mol x 169.9 g = 68.0 g AgNO 3 mol 12.5 = 68.0 g AgNO 3 100 X X = 544 g of solution

10. How much solute is present in 250. g of 5.0% K 2 CrO 4 solution? 5.0 = X 100250. g solution X = 12.5 g K 2 CrO 4

11. First find moles of each reactant molarity = moles liters molarity x liters = moles 0.642 mol x 0.0805 L = 0.0517 mol Ba(NO 3 ) 2 L 0.743 mol x 0.0445 L = 0.0331 mol KOH L 0.0517 mol 0.0331 mol 1 Ba(NO 3 ) 2 + 2 KOH → 1 Ba(OH) 2 + 2 KNO 3

12. 0.0517 mol 0.0331 mol 1 Ba(NO 3 ) 2 + 2 KOH → 1 Ba(OH) 2 + 2 KNO 3 1 Ba(NO 3 ) 2 = 0.0517 1 Ba(OH) 2 x = 0.0517 mol Ba(OH) 2 2 KOH = 0.0331 mol 1 Ba(OH) 2 x = 0.0166 mol Ba(OH) 2 --limiting reactant = KOH 0.0166 mol Ba(OH) 2 x 171.3g = 2.84g Ba(OH) 2 1 mol