CHEMISTRY 21A Instructor: Dr. Dragan Marinkovic QUIZ No. 6

solutions 4/20/2009 ______Student Name / registration number Date

1. A term miscible is used for: (1 point)

a) an ionic compound completely soluble b) mixture of oil and water c) liquids that dissolve in each other in any ratio d) a substance that dissolves to a significant extent in a solvent

2. Molarity is a solution concentration expresseed as: (1 point)

a) a number of moles of solute in 100 mL of solvent b) a number of moles of solute in 100 g of solvent c) a number of moles of solute in 100 moles of solvent d) a number of moles of solute in 1 L of solvent

3. A process in which solvent molecules, other smal molecules and hydrated ions pass from solution throug a membrane is called: (1 point)

a) dialysis b) colligative property c) osmosis d) osmotic pressure

4. Calculate molarity of 35.2 g of Ca(NO3)2 dissolved in water to make 150 mL of solution: (3 points)

-1 F.W. of Ca(NO3)2 is 40.08 + 14.01x2 + 16.00x6 = 164.10 gmol -1 number of moles of Ca(NO3)2 in 35.2 g is: 35.2 g/164.10 gmol = 0.2145 mol =0.214 mol number of moles of Ca(NO3)2 in 1 L of solurion (M = molarity) is: 0.214 mol/0.150 L = M/1 L M = 0.214/0.150 = 1.4267 M = 1.43 CHEMISTRY 21A Instructor: Dr. Dragan Marinkovic QUIZ No. 6

4a. Bonus/extra point question Why do fish sometimes die when the temperature of water they live in increases? (1 point) Solubility of oxygen (and gases in general) decreases with increase in temperature.

5. How many grams of solid calcium hydroxide will exactly react with 250 mL of 0.40 M hydrobromic acid? (4 points) Ca(OH)2(s) + 2HBr(aq) → CaBr2(aq) + 2H2O(l)

1 mol of will react exactly with 2 moles of HBr 250 mL of 0.40 M HBr contains 0.250 x 0.40 = 0.10 moles of HBr 0.10 moles of HBr wil react with exactly 0.10/2 moles of Ca(OH)2 i.e. with 0.05 moles of Ca(OH)2 F.W. of Ca(OH)2 is 40.08 + 2(1.01 + 16.00) = 74.10 0.05 moles of Ca(OH)2 is 0.05 x 74.10 = 3.705 g = 3.7 g Ca(OH)2

6. Explain how would you prepare the following dilute solutions from the more concentrated ones: (4 points)

a) 200 mL of 6% (v/v) hydrogen peroxide from 30% (v/v) hydrogen peroxide cV = c1V1 200 mL x 6% = X x 30% X = (200 mL x 6%)/30% = 40 mL The dilute solution will be made with 40 mL of 30% hydrogen peroxide and (200 – 40)160 mL of water

b) 150 mL of saline solution, 0.89% (w/v) NaCl from 6.00% (w/v) sodium chloride

cV = c1V1 150 mL x 0.89% = X x 6% X = (150 mL x 0.89%)/6% = 22.25 mL The 0.89% saline solution will be made from 22.25 mL of 6% saline solution and (150 mL – 22.25 mL) 127.75 mL of water

7. Calculate the molarity (M) of the following solutions: (3 points) a) 0.350 g of solid K2SO4 dissolved in water to make 10 mL of solution

-1 F.W. of K2SO4 is 2 x 39.10 + 32.06 + 4 x 16.00 = 174.26 gmol -1 number of moles of K2SO4 in 0.350 g is 0.350 g/174.26 gmol = 0.002008 mol = -3 2.01x10 mol K2SO4 -3 -1 the molarity of K2SO4 solution is: 2.01x10 mol/0.01 L = M/1 L M = 2.01x10 = 0.201

b) 9.50 g of solid sucrose (C12H22O11) dissolved in water to make 250 mL of solution

-1 F.W. of C12H22O11 is 12 x 12.00 + 22 x 1.01 + 11 x 16.00 = 342.22 gmol number of moles of C12H22O11 in 9.50 g is 9.50 g/342.22 = 0.02776 mol = 0.0278 mol C12H22O11

2 CHEMISTRY 21A Instructor: Dr. Dragan Marinkovic QUIZ No. 6

the molarity of C12H22O11 solution is: 0.0278 mol/0.250 L = M/1 L M = 0.1112 = 0.111

8. Calculate the concentration in %(v/v) of the following solutions: (3 points) a) 150 mL of solution contains 12.0 mL of alcohol (12 mL/150 mL) x 100 = 8%

b) The blood serum acetone level for a person was determined to be 1.8 mg of acetone per 100 mL of serum (density of acetone is 0.79 g/mL)

d = m/V V = m/d = 0.00187 g/0.79 gmL-1 = 0.0023 mL %(v/v) of acetone is (0.0023 mL/100 mL)x100 = 0.0023 % = 2.3x10-3%