APCHEM - David Kim Period 7-8 AP Chem 10/10/10 Chapter 4...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
David Kim AP Chem Period 7-8 10/10/10 Chapter 4 Page 171 Q 22,24,34,36,38,42,44,46,48,50,68,82,86,100,110,112 22. A solution of ethanol (C 2 H 5 OH) in water is prepared by dissolving 75.0 mL of ethanol (density = 0.79 g/cm 3 ) in enough water to make 250.mL of solution. What is the molarity of the ethanol in this solution? .79 g/cm 3 (1 cm 3 / 1 mL) (75.0 mL) = 59.3 g 59.3 g (1 mol/ 46.1 g) = 1.29 mol 1.29 mol / 250. mL (1000 mL / 1 L) = 5.16 M 24. Calculate the concentrations of all ions present in each of the following solutions of strong electrolytes. a. 0.0200 mol of sodium phosphate in 10.0 mL of solution. (Na 3 PO 4 ) .0200 mol / 10.0 mL (3) (1000 mL / 1L) = 6 M of Na + .0200 mol / 10.0 mL (1000 mL / 1L) = 2M of PO 4 3- b. 0.300 mol of barium nitrate in 600.0 mL of solution. Ba(NO 3 ) 2 .300 mol / 600.0 mL (1000 mL / 1L) = .5 M of Ba 2+ .300 mol / 600.0 mL (2) (1000 mL / 1L) = 1M of NO 3 1- c. 1.00g of potassium chloride in 0.500 L of solution. KCl 1.00 g (1 mol / 74.551 g)= 0.0134 mol 0.0134 mol / .500 L = .0250 M of K 0.0134 mol / .500 L = .0250 M of Cl d. 132g of ammonium sulfate in 1.50L of solution. (NH 4 ) 2 SO 4 132 g (1 mol / 132 g) = 1 mol 1 mol / 1.50 L (2) = 1.33 M of NH 4 1 mol / 1.50 L = .667 M of SO 4
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
34. A stock solution containing Mn 2+ ions was prepared by dissolving 1.584 g pure manganese metal in nitric acid and diluted to a final volume of 1.000 L. The following solutions were then prepared by dilution: 1.584 g (1 mol/ 54.9380 g) = .02883 mol of Mn 2+ .02883 mol / 1.000 L = .02883 M For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL M 1 X V 1 = M 2 X V 2 (.02883 M)(50.00 mL) = X(1000.0 mL) X = .001442 M For solution B, 10.00 mL of solution A was diluted to 250.0 mL M 1 X V 1 = M 2 X V 2 (.001442 M)(10.00 mL) = X(250.0 mL) X = .00005768 M For solution C, 10.00 mL of solution B was diluted to 500.0 mL M 1 X V 1 = M 2 X V 2 (.00005768 M)(10.00 mL) = X(500.0 mL) X = 1.154 X 10 -6 M 36. On the basis of the general solubility rules given in Table 4.1, predict which of the
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/19/2010 for the course BIO 0081 taught by Professor Haowu during the Spring '10 term at CUNY York.

Page1 / 6

APCHEM - David Kim Period 7-8 AP Chem 10/10/10 Chapter 4...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online