Chapter 13 - Solutions

What is dilution equation?

Dilution Equation M1V1 = M2V2

What is the equation of molarity?

Molarity (M) = Moles solute/Liters solution

What does a nonpolar solvent dissolve?

Nonpolar solvents dissolve nonpolar solutes. Like dissolves like.

Determine how many grams of sucrose (C12H22O11) are contained in 1.72 L of 0.758 M sucrose solution. Given: 1.72 L; 0.758 M Find: grams of sucrose

Molar Mass C12H22O11 = 342.3265 g/mol Molarity (M) = Moles solute/Liters solution .758 = Moles Solution/1.72L = .758 * 1.72 = 1.30376 moles 1.30376 mol X 342.2965 g/1 mol = 446 grams

What is osmotic pressure?

Osmotic pressure is the pressure required to stop the osmotic flow.

Calculate the freezing point (Tf) of 2.6 m sucrose solution. Given: 2.6 m Find: Freeing Point <> Tf

Tf = m X Kf Kf of water = 1.86 Tf = 2.6 X 1.86 = 4.836 Degrees C 0 Degrees C - 4.836 = -4.8 Degrees C

What is a concentrated solution?

A concentrated solution is a solution containing large amounts of solute relative to solvent.

What is a nonelectrolyte solution?

A nonelectrolyte solution is a solution containing a solute that dissolves as molecules; therefore, the solution does not conduct electricity. In general, soluble molecular solids form nonelectrolyte solutions.

What is a semipermeable membrane?

A semipermeable membrane is a membrane that allows some substances to pass through but not others.

What is stock solutions?

A stock solution is a concentrated form in which solutions are often stored.

What is the Kf of water?

The Kf of water is 1.86

What is the molarity of a solution in which 100.0 mL of 1.0 M KCl is diluted to 1.0 L a) 0.10 M b) 1.0 M c) 10.0 M

M1V1 = M2V2 M1 = 1.0 M; V1 = 0.1 L; V2 = 1.0 L M2 = M1V1/V2 M2 = (1 * .1)/1 = 0.1 M

What does a polar solvent dissolve?

Polar solvents dissolve polar or ionic solutes. Like dissolves like.

What does salt do to ice?

Salt lowers the melting point of ice.

What is a membrane?

The membranes of living cells act as semipermeable membranes.

Given: 8.5% CO2 by mass; 28.6 L solution d = 1.03g/mL Find: g CO2

8.5 g CO2 = 100 grams solution 1.03 grams = 1 mL 28.6 L X 1000 mL/1 L = 28600 mL 28600 mL X 1.03 grams/1 mL = 29,458 grams 29,458 g X 8.5 g CO2/100 g solution = 2503.93 g CO2 = 2.5 X 10^3 g CO2

What is osmosis?

Osmosis is the flow of solvent from a lower-concentration solution through a semipermeable membrane to a higher-concentration solution.

What is recrystallization?

Recrystallization is a technique used to purify a solid; involves dissolving the solid in a solvent at high temperature, creating a saturated solution, then cooling the solution to cause the crystallization of the solid. Recrystallization is a common way to purify a solid.

When temperature increases in a gas does solubility increase or decrease?

Solubility decreases in gases when temperature increases. The solubility of a gas in a liquid increases with increasing pressure.

When temperature increases in a solid does solubility increase or decrease?

Solubility increases in a solid when temperature increases.

What is solubility?

Solubility is the amount of a compound, usually in grams, that will dissolve in a certain amount of a solvent.

A soft drink contains 11.5% sucrose (C12H22O11) by mass. What volume of the soft drink solution in mL contains 85.2 g of sucrose? (Assume a density of 1.04 g/mL) Given: 11.5% sucrose by mass; 85.2 g of sucrose; d = 1.04/mL Find: Volume = mL solution (soft drink)

1.04 grams = 1 mL 11.5 g sucrose = 100 grams sucrose solution 85.2 g X 100 g sucrose sol/11.5 g sucrose = 740.869 g sucrose solution 740.869 g sucrose sol X 1 mL/1.04 grams of sol = 712.34 mL = 712 mL

How much sucrose in grams is contained in 355 mL (12 oz) of the soft drink? (Assume the density is 1.04 g/mL) 11.5% sucrose. Given: 11.5% sucrose; 1.04 grams of sol = 1 mL of soultion; 355 mL Find: sucrose in grams

1.04 grams of solution = 1 mL of solution 11.5 g sucrose = 100 grams sucrose solution 355 mL X 1.04 g/1 mL = 369.2 g of solution 369.2 g of solution X 11.5 g sucrose/100 g of solution = 42.5 g sucrose

A 25.0 mL sample of HNO3 solution requires 35.7 mL of 0.108 M Na2CO3 to completely react with all of the HNO3 in the solution. What is the concentration of the HNO3 solution? 2HNO3 (aq) + Na2CO3 (aq) ---> H2O (l) + CO2 + 2NaNO3 (aq) Given: 25.0 mL HNO3; 35.7 mL Na2CO3; 0.108 M Na2CO3 Find: HNO3 Concentration = M

2 mol HNO3 = 1 mol Na2CO3 0.108 mol Na2CO3 = 1 L Na2CO3 25.0 mL HNO3 = 0.025 L HNO3 35.7 mL Na2CO3 = 0.0357 L Na2CO3 0.0357 L Na2CO3 X 0.108 mol Na2CO3/1 L Na2CO3 = 0.0038556 mol Na2CO3 0.00386 mol Na2CO3 X 2 mol HNO3/1 mol Na2CO3 = 0.0077 mol HNO3 Molarity = 0.0077 mol HNO3/0.025 L = 0.308 M

How many milliters of 0.112 M Na2CO3 will completely react with 27.2 mL of 0.135 M HNO3 according to the reaction? 2HNO3 (aq) + Na2CO3 (aq) ---> H2O (l) + CO2 + 2NaNO3 (aq) Given: 0.112 M Na2CO3; 27.2 mL HNO3; 0.135 M HNO3 Find: mL of Na2CO3

2 mol HNO3 = 1 mol Na2CO3 0.112 mol NaCO3 = 1 L of NaCO3 0.135 mol HNO3 = 1 L of HNO3 27.2 mL = 0.0272L HNO3 0.0272 L HNO3 X 0.135 mol HNO3/1 L HNO3 = 0.003672 mol HNO3 0.003672 mol HNO3 X 1 mol Na2CO3/2 mol HNO3 = 0.001836 mol Na2CO3 0.001836 mol Na2CO3 X 1 L Na2CO3/0.112 mol Na2CO3 = .01639 L = 16.4 mL NaCO3

Consider the precipitation reaction. 2KI (aq) + Pb(NO3)2 (a) ----> PbI2 (s) + 2KNO3 (aq) How much 0.115 M KI solution will completely precipitate to Pb^2 in 0.104 L of 0.225 M Pb(NO3)2. Given: 0.115 M KI; 0.104 L Pb(NO3)2 solution; 0.225 M Pb(NO3)2 Find: L KI Solution

2 mol KI = 1 mol Pb(NO3)2 Molarity (M) = Moles solute/Liters Solution Molarity = mol of solution/1 liter of solution Molarity (M) = 0.225 mol Pb(NO3)2/1L Pb(NO3)2 0.104 L Pb(NO3)2 X 0.225 mol Pb(NO3)2/1 L Pb(NO3)2 = 0.0234 mol Pb(NO3)2 0.0234 mol Pb(NO3)2 X 2 mol KI/1 mol Pb(NO3)2 = 0.0468 mol KI 0.0468 mol KI X 1 L KI/0.115 KI = 0.407 L KI solution

Ocean water contains 3.5% NaCl by mass. How much salt can be obtained from 254 g of seawater.

254 g X 3.5 g/100 g = 8.89 g NaCl

What is the equation for the boiling point elevation (Tb) of a solution?

<>Tb = m X Kb Tb = Change in temperature of the boiling point in degrees C m = molarity of the solution = mol solute/kg solvent Kb = Boling point Elevation constant for the solvent Kb water = 0.512

Calculate the freezing point of a 1.7 m ethylene glycol solution. Given: 1.7 m Find: Freezing Point <>Tf

<>Tf = m X Kf Kf of water = 1.86 Tf = 1.7 X 1.86 = 3.2 Degrees C Freeing Point = 0 - 3.2 = -3.2 degrees C

What is freezing point depression?

Freezing point depression is the decrease in the freeing point of a solvent caused by the presence of a solute.

What is the solute and solvent phase of a gaseous solution?

Gaseous Solution: Solute Phase - Gas Solvent Phase - Gas Example - Air

What is Henry's Law?

Henry's Law: The solubility of gases depends on pressure. The higher the pressure above a liquid, the more soluble the gas is in the liquid.

What is molarity?

Molarity is a common unit of solution concentration, defined as the number of moles of solute per liter of solution.

Calculate the freezing point (Tf) of a water solution at each concentration. a) 0.85 m b) 1.45 m c) 4.8 m d) 2.35 m

Tf = m X Kf Kf of water = 1.86 a) 0.85 m Tf = 0.85 X 1.86 = 1.581 Degrees C 0-1.581 = -1.6 Degrees C b) 1.45 m Tf = 1.45 X 1.86 = 2.697 Degrees C 0-2.697 = -2.70 Degrees C c) 4.8 m Tf = 4.8 X 1.86 = 8.928 Degrees C 0-8.928 = -8.9 Degrees C d) 2.35 m Tf = 2.35 X 1.86 = 4.371 Degrees C 0-4.371 = -4.37 Degrees C

What is the Kb of water?

The Kb of water is 0.512

What is the solubility of solids in water highly dependent on?

The solubility of solids in water can be highly dependent on temperature. In general, the solubility of solids in water increases with increasing temperature.

Calculate the molarity of each solution. a) 0.127 mole of sucrose in 655 mL of solution b) 0.205 mole of KNO3 in 0.875L of solution c) 1.1 mol of KCl in 2.7 L of solutoin

a) 0.127 mole of sucrose in 655 mL of solution 655 mL = .655 L .127/.655 = .194 M b) 0.205 mole of KNO3 in 0.875L of solution .205/.875 = .234 M c) 1.1 mol of KCl in 2.7 L of solutoin 1.1/2.7 = .41 M

What is the solute and solvent phase of a solid solution?

Solid Solution: Solute Phase - Solid Solvent Phase - Solid Example - Brass

What is a solute?

A solute is the minority component of a solution.

A solution is saturated in both nitrogen gas (N2) and potassium chloride (KCl) at 75 degrees C. What happens when the solution is cooled to room temperature? a) Some nitrogen gas bubbles out of the solution b) Some potassium chloride precipitates out of the solution. c) Both a and b d) Nothing happens

b) Some potassium chloride preciptiates out of the solution. The solubility of most solids decreases with decreasing temperature. However, the solubility of gases increases with decreasing temperature. Therefore, the nitrogen becomes more soluble and will not bubble out of the solution.

Which solution has the highest boiling point? a) 0.50 m C12H22O11 b) 0.50 m C6H12O6 b) 0.50 m C2H6O2 d) All of these solutions will have the same boiling point

d) All of these solutions will have the same boiling point Since the boiling point elevation depends only on the concentration of the dissolved particles, and not on the kind of dissolved particles, all of these solutions have the same boiling point.

What is the equation for the freezing point depression of a solution?

freezing point depression of a solution = Change of Temp of the freeing point in C = m (molarity) X K (the freezing point depression constant for the solvent) <>Tf = m X Kf <> T = the change in temperature of the freezing point in degrees C m = the molarity of the solution in mol solute/kg solvent Kf = the freezing point depression constant for the solvent

How much 0.125 M NaOH solution do we need to completely neutralize 0.225 L of 0.175 M H2SO4 solution? Given: 0.125 M NaOH; 0.225 L H2SO4; 0.175 M H2SO4 Find: L NaOH H2SO4 + 2NaOH = Na2SO4 + 2H2O

2 mol NaOH = 1 mole

What is a dilute solution?

A dilute solution is a solution containing small amounts of solute relative to solvent.

What is a saturated solution?

A saturated solution is a solution that holds the maximum amount of solute under the solution conditions. If additional solute is added to a saturated solution, it will not dissolve.

What is a solution?

A solution is a homogeneous mixture of two or more substances. Solutions are common - Most of the liquids and gases we encounter every day are actually solutions. A solution has at least two components.

What are solutions composed of?

A solution may be composed of a gas and a liquid, a liquid and another liquid, a solid and a gas, or other combinations.

What is a solvent?

A solvent is the majority component of a solution. Water is a common solvent.

What is a supersaturated solution?

A supersaturated solution is a solution holding more than the normal maximum amount of solute. The solute will normally precipitate from (or come out of) a supersaturated solution.

What is an aqueous solution?

An aqueous solution is the most common solution containing a solid, a liquid, or a gas and water. They are critical to life. Examples - Sugar Water, Salt water

What is an electrolyte solution?

An electrolyte solution is a solution containing a solute that dissociates into ions. In general, soluble ionic solids form electrolyte solutions.

What is an unsaturated solution?

An unsaturated solution is a solution holding less than the maximum possible amount of solute under the solution conditions. If additional solute is added to an unsaturated solution, it will dissolve.

What is the boiling point elevation?

Boiling point elevation the increase in the boiling point of a solution caused by the presence of the solute.

Which compound would you expect to be least soluble in water? a) CCl4 b) CH3Cl c) NH3 d) KF

CH3CL and NH3 are both polar compounds and KF is ionic. All three would therefore interact more strongly with the water molecules (which are polar) than CCl4 which is nonpolar. a) CCl4

Determine the molar concentration of Ca2+ and Cl- in a 0.75 M of CaCl2 solution. Given: 0.75 M Find: Molarity of Ca2+ and Cl-

CaCl2 Molarity of Ca2+ = 1(.75 M) = .75 M Molarity of Cl- = 2(.75 M) = 1.5 M

What are colligative properties?

Colligative properties are the physical properties of solutions that depend on the number of solute particles present but not the type of solute particles.

Consider the following reaction occurring in aqueous solution. A (aq) + 2B (aq) ---> Products What volume of a 0.100 M solution of B is required to completely react with 50.0 mL of a 0.200 M solution A? a) 25.0 mL b) 50.0 mL c) 100.0 mL d) 200.0 mL

Given: 0.100 M B; 50.0 ml A; 0.200 M A Find: B mL 1 mol A = 2 mol B 0.1 mol B = 1 L B 0.2 mol A = 1 L A 50.0 mL A = 0.05 L A 0.05 L A X 0.2 mol A/1 L A = 0.01 mol A 0.01 mol A X 2 mol B/1 mol A = 0.02 mol B 0.02 mol B X 1 L B/.1 mol B = 0.2 L B = 200 mL B d) 200.0 mL

What is the solute and solvent phase of a liquid solution?

Liquid Solution: Solute Phase - Gas Solvent Phase - Liquid Example - Soda Water Liquid Solution: Solute Phase - Liquid Solvent Phase - Liquid Example - Vodka Liquid Solution: Solute Phase - Solid Solvent Phase - Liquid Example - Seawater

A 122 mL sample of a 1.2 M sucrose solution is diluted to 500.0 mL. What is the molarity of the diluted solution. Given: V1 = 122 mL; V2 = 500 mL; M1 = 1.2 M Find: M2

M1V1 = M2V2 M2 = M1V1/V2 V1 = 122 mL ; V2 -= 500 mL M2 = (1.2 * 122)/500 = .29 M

A lab procedure calls for 5.00 L of a 1.50 M Cl solution. How should we prepare this solution from a 12.0 M stock solution? Given: M1 = 12.0 M; M2 = 1.5 M; V2 = 5.00 L Find: V1

M1V1 = M2V2 V1 = (M2V2)/M1 V1 = (1.5 * 5.00) / 12.0 = 0.625 L

How much 6.0 M NaNO3 solution should you use to make 0.585 L of a 1.2 M NaNO3 solution? Given: M1 = 6.0 M; M2 = 1.2 M; V2 = 0.585 L Find: V1

M1V1 = M2V2 V1 = M2V2/M1 V1 = (1.2 * 0.585)/6 = 0.117 = 0.12 L

To what volume should you dilute 0.100 L of a 15 M NaOH solution to obtain a 1.0 M NaOH solution? Given: V1 = 0.100 L; M1 = 15 M; M2 - 1.0 M Find: V2

M1V1 = M2V2 V2 = M1V1/M2 V2 = (15 * 0.100)/1 = 1.5 L

Calculate the mass percent of a solution containing 27.5 grams of ethonol (C2H6O) and 175 mL of H2O. (Assume the density of water is 1.00 g/mL) Given: 27.5 g C2H6O; 175 mL H2O Find: Mass Pecent

Mass Percent = Mass Solute/(Mass Solute + Mass Solvent) X 100% 1 g H2O = 1 mL = 175 mL = 175 g H2O Mass Percent = 27.5 g / (27.5 g + 175 g) X 100% = 13.6% The solution is 13.6% C2H6O by mass.

Calculate the mass percent of a sucrose solution containing 11.3 grams of sucrose and 412.1 mL of water. (Assume that the density of water is 1.00 g/mL)

Mass Percent = Mass Solute/(Mass Solute + Mass Solvent) X 100% 1 g H2O = 1 mL H2O = 412.1 mL H2O = 412.1 g H2O Mass Percent = 11.3g / (11.3 g + 412.1 g) X 100% = 2.67% The solution is 2.67% sucrose by mass.

Calculate the mass percent of NaCl in a solution containing 15.3 grams of NaCl and 155.0 grams of water. Given: 15.3 grams NaCl; 155.0 grams H2O Find: Mass Percent

Mass Percent = Mass Solute/(Mass Solute + Mass Solvent) X 100% Mass Percent = 15.3 g / (15.3 g + 155.0 g) X 100% = 8.98% The solution is 8.98% NaCl by mass.

Calculate the concentration of each solution in mass percent. a) 41.2 g C12H22O11 in 498 g H2O b) 178 mg C6H12O6 in 4.91 g H2O c) 7.55 g NaCl in 155 g H2O

Mass Percent = Mass Solute/(Mass Solute + Mass Solvent) X 100% a) 41.2 g C12H22O11 in 498 g H2O Mass Percent = 41.2 g/(41.2 + 498) X 100% = 7.64% b) 178 mg C6H12O6 in 4.91 g H2O 178 mg X 1 g/1000 mg = .178 g Mass Percent = .178 / (.178 + 4.91) X 100% = 3.50% c) 7.55 g NaCl in 155 g H2O Mass Percent = 7.55 / (7.55 + 155) X 100% = 4.64%

What is mass percent?

Mass percent (or mass percent composition) is the percentage by mass of each element in a compound. A common method of reporting solution concentration is mass percent. Mass percent is the number of grams of solute per 100 grams of solution. Mass Percent = Mass Solute/(Mass Solute + Mass Solvent) X 100%

How much of a 0.225 M KCl solution contains 55.8 grams of KCl. Given: 0.225 M; 55.8 grams Find: Liters

Molar Mass KCl = 74.5513 g/mol Molarity (M) = Moles solute/Liters solution 55.8 g X 1 mol KCl/74.5513 = 0.7484 mol KCl 0.225 M = 0.7484 mol/Liters = 0.7484 mol/0.225 = 3.33 L

Calculate the molarity of a solution made by putting 55.8 g of NaNO3 into a beaker and diluting to 2.50 L.

Molar Mass of NaNO3 = 84.9947 g/mol 1 mol NaNO3 = 84.9947 g 55.8 g X 1 mol NaNO3/84.9947 = 0.6565 mol NaNO3 Molarity = 0.6565/2.5 = 0.263 M

How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH? Given: 0.114 M NaOH; 1.24 mol NaOH Find: L solution

Molarity (M) = Moles solute/Liters solution .114 M = 1.24 moles/Liters = 1.24 moles/.114 M = 10.9 L

Caclulate the molarity of sucrose solution made with 1.58 moles of sucrose diluted to a total volume of 5.0 L of solution. Given: 1.58 mol of sucrose; 5.0 L of solution Find: Molarity (M)

Molarity (M) = Moles solute/Liters solution Molarity = 1.58/5.0 = 0.32 M

What volume of each solution contains 0.15 mol of KCl? a) 0.255 M KCl b) 1.8 M KCl c) 0.995 M KCl

Molarity (M) = Moles solute/Liters solution a) 0.255 M KCl 0.225 M = 0.15 moles/Liters = 0.15 moles/.255 M = .59 Liters b) 1.8 M KCl 1.8 M = 0.15 moles/Liters = 0.15/1.8 = .083 Liters c) 0.995 M KCl 0.995 M = 0.15 moles/Liters = 0.15/.995 = 0.15 Liters

Calculate the molarity of a solution made by putting 15.5 grams NaCl into a beaker and adding water to make 1.50 L of NaCl solution. Given: 15.5 g NaCl; 1.50 L NaCl Find: Molarity M

Molarity (M) = Moles solute/Liters solution Molar Mass of NaCl = 58.44 g 1 mol NaCl = 58.44 g 15.5g X 1 mol NaCl/58.44 g = .2652 mol NaCl Molarity (M) = Moles solute/Liters solution Molarity (M) = 0.2652 mol NaCl/1.5 L = 0.177 M

How many moles of NaCl are contained in each solution? a) 1.5 L of a 1.2 M NaCl solution b) 0.448 L of a 0.85 M NaCl solution c) 144 mL of a 1.65 M NaCl solution

Molarity (M) = Moles solute/Liters solution Molar Mass of NaCl = 58.4428 g/mol a) 1.5 L of a 1.2 M NaCl solution 1.2 M = moles/1.5 L = 1.8 mol NaCl b) 0.448 L of a 0.85 M NaCl solution 0.85 M = moles/0.448 L = 0.38 mol NaCl c) 144 mL of a 1.65 M NaCl solution 1.65 M = moles/.144 L = .238 mol NaCl

Calculate the molarity of each solution. a) 22.6 grams of C12H22O11 in 0.442 L of solution b) 42.6 grams of NaCl in 1.58 L of solution c) 315 mg of C6H12O6 in 58.2 mL of solution

Molarity (M) = Moles solute/Liters solution a) 22.6 grams of C12H22O11 in 0.442 L of solution Molar Mass C12H22O11 = 342.2965 g/mol 22.6 g X 1 mol /342.2965 = .066 mol Molarity = .066/.442 = 17502 M = .149 M b) 42.6 grams of NaCl in 1.58 L of solution Molar Mass of NaCl = 58.4428 g/mol 42.6 g X 1 mol/58.4428 = .7289 mol Molarity = .7289/1.58 = .461 M c) 315 mg of C6H12O6 in 58.2 mL of solution Molar Mass of C6H12O6 = 180.1559 g/mol 315 mg = .315 grams 58.2 mL = .0582 L .315 g X 1 mol/180.1559 = .0017 Molarity = .0017/.0582 = 0.0300 M

Calculate the molarity of each solution. a) 0.25 mol solute; 0.250 kg solvent b) 0.882 mol solute; 0.225 kg solvent c) 0.012 mol solute; 23.1 g solvent

Molarity (m) = Moles Solute/Kilograms Solvent a) 0.25 mol solute; 0.250 kg solvent Molarity (m) = 0.25/0.250 = 1 m b) 0.882 mol solute; 0.225 kg solvent Molarity (m) = 0.882/0.225 = 3.92 m c) 0.012 mol solute; 23.1 g solvent 23.1 grams = 0.0231 kg Molarity = 0.012/.0231 = .52 m

Calculate the molarity of a solution containing 12.5 grams of ethylene glycol (C2H6O2) dissolved in 135 grams of water. Given: 12.5 g C2H6O2; 135 g H2O Find: Molarity (m)

Molarity (m) = Moles Solute/Kilograms Solvent Molar Mass C2H6O2 = 62.0678 g/mol 12.5 g X 1 mol/62.0678 = .20139 mol 135 g H2O = .135 kg Molarity (m) = .20139 / .135 = 1.49 m

Calculate the molarity (m) of a sucrose (C12H22O11) solution containing 50.4 grams and 0.332 kg of water. Given: 50.4 g C12H22O11; 0.332 kg H2O Find: molarity (m)

Molarity = Moles Solute/Kilograms Solvent Molar Mass = 342.2965 g/mol 50.4 g C12H22O11 X 1 mol/342.2965 = 0.1472 mol C12H22O11 Molarity (m) = 0.1472 mol/0.332 kg H2O = 0.443 m

Calculate the molarity of a solution containing 17.2 grams of ethylene glycol (C2H6O2) dissolved in 0.500 kg of water. Given: 17.2 g C2H6O2; 0.500 kg H2O) Find: Molarity of Solution

Molarity = Moles Solute/Kilograms Solvent Molar Mass of C2H6O2 = 62.0678 g/mol 17.2 g X 1 mol/62.0678 = .27711 moles C2H6O2 Molarity = .27711 moles/0.500 kg = .554 m

A solution is 0.15 M in K2SO4. What is the concentration of K+ in solution? a) 0.075 M b) 0.15 M c) 0.30 M d) 0.45 M

Molarity of K+ = 2(.15 M) = 0.30 M C

Determine the molar concentration of Na+ and PO4^-3 in a 1.50 M Na3PO4 solution. Given: 1.5 M Find: Molarity (M) of Na+ and PO4^-3

Na3PO4 Molarity of Na+ = 3(1.5 M) = 4.5 M Molarity of PO4^-3 = 1(1.50 M) = 1.5 M

Calculate the boiling point of a 1.7 m ethylene glycol solution. Given: 1.7 m Find: Tb

Tb = m X Kb Kb = 0.512 Tb = 1.7 X 0.512 = 0.8704 Degrees C 100 Degrees C + 0.87 = 100.87 Degrees C

Calculate the boiling point of a 3.5 m glucose solution. Given: 3.5 m Find: Tb

Tb = m X Kb Kb = 0.512 Tb = 3.5 X 0.512 = 1.792 100 + 1.8 = 101.8 Degrees C

Calculate the boiling point of a water solution at each concentration. a) 0.118 m b) 1.94 m c) 3.88 m d) 2.16 m

Tb = m X Kb Kb = 0.512 a) 0.118 m Tb = 0.118 m X 0.512 = 0.060416 100 + 0.0604 = 100.060 Degrees C b) 1.94 m Tb = 1.94 X 0.512 = 0.99328 100 + 0.993 = 100.993 Degrees C c) 3.88 m Tb = 3.88 X 0.512 = 1.98656 100 + 1.99 = 101.99 Degrees C d) 2.16 m Tb = 2.16 X 0.512 = 1.10592 100 + 1.11 = 101.11 Degrees C

Determine the concentration of Cl- in each aqueous solution. a) 0.15 M NaCl b) 0.15 M CuCl2 c) 0.15 M AlCl3

a) 0.15 M NaCl Molarity of Cl- = 1(0.15 M) = 0.15 M Cl- b) 0.15 M CuCl2 Molarity of Cl- = 2(0.15 M) = 0.30 M Cl- c) 0.15 M AlCl3 Molarity of Cl- = 3(0.15 M) = 0.45 M Cl-

A lab procedure calls for a 2.0 molal aqueous solution. A student accidentally makes a 2.0 molar solution. The solution made by the student is: a) too concentrated b) too dilute c) just right d) it depends on the molar mass of the solute

a) too concentrated A 2.0 m solution would be made by adding 2 mol of soulute to 1 kg of solvent. 1 kg of water has a volume of 1 L, but because of the dissolved solute, the final solution would have a volume of slightly more than 1 L. A 2.0 M solution, by contrast would consist of 2 mol of solute in a solution of exactly 1 L. Therefore a 2 M aqueous solution would be slightly more concentrated than a 2 m solution.