percent (%) concentration % (w/v) concentration:

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Percent (%) concentration % (w/v) concentration: mass of solute in grams contained by 100 mL solution % (w/w) concentration: mass of solute in grams contained in 100 g of solution % (v/v) concentration: volume of solute in mL in 100 mL solution

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Percent (%) concentration % (w/v) concentration: mass of solute in grams contained by 100 mL solution % (w/w) concentration: mass of solute in grams contained in 100 g of solution % (v/v) concentration: volume of solute in mL in 100 mL solution. - PowerPoint PPT Presentation

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Page 1: Percent (%) concentration % (w/v) concentration:

Percent (%) concentration

% (w/v) concentration: mass of solute in grams contained by 100 mL solution

% (w/w) concentration: mass of solute in grams contained in 100 g of solution

% (v/v) concentration: volume of solute in mL in 100 mL solution

Page 2: Percent (%) concentration % (w/v) concentration:

parts per million (ppm) are used to express the concentrations of very dilute solution

A solution whose solute concentration is 1 ppm contains 1 g of solute for each million (106) grams of solution or, equivalently, 1 mg of solute per kilogram of solution

A 2.5-g sample of groundwater was found to contain 5.4 g of Zn2+. What is the concentration of Zn2+ in parts per million?

Page 3: Percent (%) concentration % (w/v) concentration:

Mole Fraction, Molarity, and Molality are concentration expressions based on the number of moles of one or more components of the solution The symbol X X is used for mole fractionmole fraction, with a subscript to indicate the component of interest. The mole fraction of HCl is represented as XHCl. The sum of the mole fractions of all components of The sum of the mole fractions of all components of a solution must equal 1 a solution must equal 1

A solution of hydrochloric acid contains 36 percent HCl by mass. Calculate the mole fraction of HCl in the solution.

Page 4: Percent (%) concentration % (w/v) concentration:

Molarity of a solute in a solution is defined as the number of moles of solute per liters of solution

A 2.00L solution of hydrochloric acid contains 36 g of HCl. Calculate the molarity of HCl in the solution.

MHCl = ----------------0.99 mol2.00 L

= 0.495 = 0.50 M

Page 5: Percent (%) concentration % (w/v) concentration:

The molality of a solution, denoted m, is defined as the number of moles of solute per kilogram of solvent

A solution of hydrochloric acid contains 36 percent HCl by mass. Calculate the molality of HCl in the solution. (c) What additional information would you need to calculate the molarity of the solution?

Mass of water = 100 gSolu – 36 g HCl = 64 g

Page 6: Percent (%) concentration % (w/v) concentration:

(a) Calculate the mole fraction of NaOCl in a commercial bleach solution containing 3.62

mass percent NaOCl in water.

gsolution

gNaOCl

100

62.3

gsolution

gNaOCl

100

62.3

gWategNOCl

gNaOCl

38.9662.3

62.3

OgHOmolH

gWategNaOCl

molNaOClgNOCl

gNaOClmolNaOCl

gNaOCl

2

2

181

38.9644.74

162.3

44.741

62.3

molWatemolNOCl

molNaOCl

3544.5048628.0

048628.0

=0.00900

0.009000149

Page 7: Percent (%) concentration % (w/v) concentration:

(b) What is the molality of a solution made by dissolving 36.5 g of naphthalene, C10H8, in 420 g of

toluene, C7H8? (b) 0.678 m

gkg

gSolvent

HgCHmolC

HgC

m

3

810

810810

101

420

17.1281

5.36

=0.28477 / .420

0.678024

Page 8: Percent (%) concentration % (w/v) concentration:

Given that the density of a solution of 5.0 g of toluene and 225 g of benzene is 0.876 g/mL, calculate (a) the molarity of the solution; (b) the mass percentage of solute.

Page 9: Percent (%) concentration % (w/v) concentration:

Calculate the molality of a solution that contains 25 g of H2SO4 dissolved in 80. g of

H2O

Page 10: Percent (%) concentration % (w/v) concentration:

Calculate the molality of a 10.0% H3PO4 solution in water

Page 11: Percent (%) concentration % (w/v) concentration:

Calculate the molality of a solution that contains 51.2 g of naphthalene, C10H—, in 500. mL of

carbon tetrachloride. The density of CCl4 is 1.60

g/mL

Page 12: Percent (%) concentration % (w/v) concentration:

If 8.32 grams of methanol, CH3OH, are

dissolved in 10.3 grams of water, what is the mole fraction of methanol in the solution?

Page 13: Percent (%) concentration % (w/v) concentration:

What is the molarity of 2500. mL of a solution that contains 160. grams of NH4NO3?

Page 14: Percent (%) concentration % (w/v) concentration:

11-1Acid and base Reactions100. mL of 0.100 M HCl solution and 100. mL of 0.100 M NaOH are mixed. What is the molarity of the salt in the resulting solution? Assume that the volumes are additives. 1. Balance Chemical Equation1. Balance Chemical Equation

HCl + NaOH H2O + NaClReaction Ratio: 1 mol 1 mol 1 mol 1 mol

Start: Change:

For HCl: (100 mL)(0.100mol/L) = 10.0 mmmol HCl

10.0 m mol

For NaOH = (100. mL)(0.100mol/L) = 10.0 mmol

10.0 m mol 00

For NaCl: moles of NaCl at the end of reaction = moles of HCl at beginning = 10.0 m mol

-10.0-10.0After reaction: 0.00 0.00 10.0

MNaCl = ----------------------------10.0 mmmol NaCl(100+100) mmL solution

= 0.0500

10.010.0

Page 15: Percent (%) concentration % (w/v) concentration:

11-1Acid and base Reactions80. mL of 0.100 M HCl solution and 100. mL of 0.100 M NaOH are mixed. What is the molarity of the salt in the resulting solution? Assume that the volumes are additives. 1. Balance Chemical Equation1. Balance Chemical Equation

HCl + NaOH H2O + NaClReaction Ratio: 1 mol 1 mol 1 mol 1 molStart: Change:

For HCl: (80 mL)(0.100mol/L) = 8.0 mmmol HCl

8.0 m mol

For NaOH = (100. mL)(0.100mol/L) = 10.0 mmol

10.0 m mol 00

For NaCl: moles of NaCl at the end of reaction = moles of HCl at beginning = 8.0 m mol

-8.0-8.0After reaction: 0.00 2.0 8.0

MNaCl = ----------------------------8.0 mmmol NaCl(100+80) mmL sol

= 0.044

MNaOH = ------------------------2.0 mmol NaOH180 mL solution

= 0.011

8.0 m mol

Page 16: Percent (%) concentration % (w/v) concentration:

100. mL of 0.100 M HCl solution and 100. mL of 0.80 M NaOH are mixed. What is the molarity of the salt in the resulting solution? Assume that the volumes are additives. 1. Balance Chemical Equation1. Balance Chemical Equation

HCl + NaOH H2O + NaClReaction Ratio: 1 mol 1 mol 1 mol 1 molStart: Change:

For HCl: (100 mL)(0.100mol/L) = 10.0 mmmol HCl

10.0 m mol

For NaOH = (100. mL)(0.80mol/L) = 8.0 mmol

8.0 m mol 00

For NaCl: moles of NaCl at the end of reaction = moles of NaOH at beginning = 8.0 m mol

-8.0-8.0After reaction: 2.0 0.0 8.0

+8.0

MNaCl = ----------------------------8.0 mmmol NaCl(100+100) mmL solution

= 0.040

MHCl = ----------------------------2.0 mmmol NaCl(100+100) mmL solution

= 0.010

Page 17: Percent (%) concentration % (w/v) concentration:

100. mL of 0.100 M HCl solution and 100. mL of 0.80 M NaOH are mixed. What is the molarity of the salt in the resulting solution? Assume that the volumes are additives. 1. Balance Chemical Equation1. Balance Chemical Equation

HCl + NaOH H2O + NaClReaction Ratio: 1 mol 1 mol 1 mol 1 molStart: Change:

For HCl: (100 mL)(0.100mol/L) = 10.0 mmmol HCl

10.0 m mol

For NaOH = (100. mL)(0.80mol/L) = 80. mmol

80. m mol 00

For NaCl: moles of NaCl at the end of reaction = moles of NaOH at beginning = 10. m mol

-10.0-10.0After reaction: 0.00 70. 8.0

+10.0

MNaCl = ----------------------------10. mmmol NaCl(100+100) mmL solution

= 0.050

MNaOH = ----------------------------70. mmmol NaCl(100+100) mmL solution

= 0.035

Page 18: Percent (%) concentration % (w/v) concentration:

100. mL of 1.00 M H2SO4 solution is mixed with 200. mL of 1.00 M KOH. What is the molarity of the salt in the resulting solution? Assume that the volumes are additives. 1. Balance Chemical Equation1. Balance Chemical Equation

H2SO4 + 2KOH 2H2O + K2SO4Reaction Ratio: 1 mol 2 mol 2 mol 1 molStart: Change:

Moles of H2SO4: (100 mL)(1.00mol/L) = 100 mmmol

100 m mol

mols KOH = (200. mL)(1.00mol/L) = 200 mmol

200 m mol 00

moles of K2SO4 at the end of reaction = moles of H2SO4 at beginning = 100 m mol

-200-100After reaction: 0.00 0.0 100

M H2SO4 = ---------------------100 mmmol K2SO4

(100+200) mmL sol= 0.333

100 m mol

Page 19: Percent (%) concentration % (w/v) concentration:

TitrationTitrationTitration is a process in which a solution of one reactant, titrant, is carfully added