14. Solutions

Henry's Law Calculations

Henry's Law With Example Problem

Solutions: Crash Course Chemistry #27

6.5 &#215; 10&#8722;4 M

The solubility of a gas changes in direct proportion to the pressure of the gas over the solution.  Set up a ratio of pressures to calculate the solubility at the new pressure.

2.6 &#215; 10&#8722;3 M

Correct answer in accordance with Henry's law.

4.6 &#215; 10&#8722;3 M

1.3 &#215; 10&#8722;3 M

3.2 &#215; 10&#8722;2 M

Use the solubility of O2 at 25 &#176;C to calculate the Henry&#8217;s law constant for O2. Then use this value to determine the solubility of O2 at 2.0 atm.

At 25 &#176;C, the solubility of O2 is 4.55 &#215; 10&#8722;3 M at 3.5 atm pressure. What is the solubility of O2 at the same temperature and at 2.0 atm pressure?

The presence of the radioactive gas radon (Rn) in well water presents a possible health hazard in parts of the United
States. (a) Assuming that the solubility of radon in water
with 1 atm pressure of the gas over the water at 30 °C is
7.27 * 10-3 M, what is the Henry’s law constant for radon in water at this temperature?

A gas has a Henry’s law constant of 0.112 M>atm. What total 
volume of solution is needed to completely dissolve 1.65 L of 
the gas at a pressure of 725 torr and a temperature of 25 °C?

Henry's Law Calculations Example 2

The mole fraction of ethanol rather than glucose should be used to calculate the vapor pressure.

With a nonvolatile solute the vapor pressure of the solution will be less than the pure solvent in direct proportion to the mole fraction of the solvent.

Correct, the mole fraction of ethanol in the solution is 0.988.

To three significant figures this is the vapor pressure of ethanol not a solution of glucose in ethanol.

 Raoult&#8217;s law states that the vapor pressure of the solution is equal to the mole fraction of the solvent times the vapor pressure of the pure solvent. 

What would be the vapor pressure at 25.0 &#176;C of a solution of 5.00 g of glucose (C6H12O6) in 100.0 g of ethanol (C2H5OH)? Pure ethanol has a vapor pressure of 54.68 mmHg at 25 &#176;C.

Henry’s Law Constant for nitrogen in water is 1.67 × 10-4 M • atm–1. If a closed canister contains 0.103 M nitrogen, what would be its pressure in atm?

The Henry's law constant for CO2 in water at 25 °C is 3.1x10^-2 M atm-1. (a) What is the soubility of CO2 in water at this temperature if the soltuion is in contact with air at normal atmospheric pressure?

Henry's Law

At ordinary body temperature 137 °C2, the solubility of
N2 in water at ordinary atmospheric pressure (1.0 atm) is
0.015 g>L. Air is approximately 78 mol % N2. (b) At a depth
of 100 ft in water, the external pressure is 4.0 atm. What
is the solubility of N2 from air in blood at this pressure?

Henry's Law Calculations Concept 1

Henry's Law Calculations Concept 2

Most fish need at least 4 ppm dissolved O2 in water for survival. (b) What partial pressure of O2 above water is needed to obtain
4 ppm O2 in water at 10 °C? (The Henry’s law constant for
O2 at this temperature is 1.71 * 10-3 mol>L@atm.)

A saturated solution forms when 0.0537 L of argon, at a pressure 
of 1.0 atm and temperature of 25 °C, is dissolved in 1.0 L of 
water. Calculate the Henry’s law constant for argon.

Henry's Law Explained - Gas Solubility & Partial Pressure - Chemistry Problems

A scuba diver is breathing compressed air that is 21% (by volume) oxygen at a depth of 66 ft. beneath the surface. The total pressure at this depth is 3.0 atm. The Henry’s law constant (k) for O2 in water at a normal body temperature (37 °C ) is 1.93 * 10-3 mol>(L atm). What is the solubility
of O2 in the blood of the diver at 66 ft. beneath the surface?
(LO 13.9)
(a) 1.22 * 10-3 M (b) 2.76 * 10-2 M
(c) 1.22 * 10-1 M (d) 2.76 * 10-1 M

At an altitude of 10,000 ft, the partial pressure of oxygen in the lungs is about 68 mm Hg. What is the concentration in mg/L of dissolved O2 in blood (or water) at this partial pres- sure and a normal body temperature of 37 °C? The solubil- ity of O2 in water at 37 °C and 1 atm partial pressure is 1.93 * 10-3 mol>L.

At 0°C and 1.00 atm, as much as 0.84 g of O2 can dissolve in 1.0 L of water. At 0°C and 4.00 atm, how many grams of O2 dissolve in 1.0 L of water?

Fish generally need an O2 concentration in water of at least 4 mg/L for survival. What partial pressure of oxygen above the water in atmospheres at 0 °C is needed to obtain this concentration? The solubility of O2 in water at 0 °C and 1 atm partial pressure is 2.21 * 10-3 mol>L.

Henry's Law: Relationship Between Pressure and Solubility of a Gas

The atmospheric pressure in a lab is calculated as 1.3 atm. If oxygen gas contributes 62% of this atmospheric pressure, determine its mass (in g) dissolved at room temperature in 25 L of water. The Henry’s Law Constant for oxygen in water at this temperature is 5.3 × 10–5 M/atm.

Henry's Law: KH = P/C, Pressure and Concentration + 2 Examples

Henry's Law Calculations Example 1

The 2 Point Form of Henry's Law Formula is used when dealing with 2 pressures and 2 solubilities of a given gas.

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1. Intro to General Chemistry

2. Atoms & Elements

3. Chemical Reactions

4. BONUS: Lab Techniques and Procedures

5. BONUS: Mathematical Operations and Functions

6. Chemical Quantities & Aqueous Reactions

7. Gases

8. Thermochemistry

9. Quantum Mechanics

10. Periodic Properties of the Elements

11. Bonding & Molecular Structure

12. Molecular Shapes & Valence Bond Theory

13. Liquids, Solids & Intermolecular Forces

15. Chemical Kinetics

16. Chemical Equilibrium

17. Acid and Base Equilibrium

18. Aqueous Equilibrium

19. Chemical Thermodynamics

20. Electrochemistry

21. Nuclear Chemistry

22. Organic Chemistry

23. Chemistry of the Nonmetals

24. Transition Metals and Coordination Compounds

[BOOK CHEM] Clutch Version

Types of Aqueous Solutions

When an ionic solid dissolves, ions leave the solid and become dispersed in the solvent. Three possible aqueous solutions can be formed:saturated, unsaturated and supersaturated.

Boiling Point Elevation

Boiling Point Elevation is the phenomenon when adding a solute to a pure solvent results in increased boiling point of the solvent.

Intro to Henry's Law

According to&nbsp;Henry's Law, at a constant temperature the amount of a gas dissolved in a solvent is directly proportional to its partial pressure when it is in equilibrium with its liquid phase.&nbsp;

Osmotic Pressure

Osmotic Pressure is the force that drives Osmosis from higher concentration to lower concentration.

Mole Fraction

Mole Fraction (X) relates the moles of solute and solvent within a solution.&nbsp;

The Colligative Properties

The&nbsp;Colligative Properties&nbsp;explain what happens to a pure solvent as it transitions to a solution.

Molality

Molality (m)&nbsp;represents the number of moles of solute dissolved per kilogram of solvent.&nbsp;

Freezing Point Depression

Freezing Point Depression is the phenomenon when adding a solute to a pure solvent results in decreased freezing point of the solvent.

Solutions: Solubility and Intermolecular Forces

Solutions are homogeneous mixtures that result from the dissolving of a solute by a solvent.

Vapor Pressure Lowering (Raoult's Law)

Vapor Pressure is the pressure exerted by a gas at equilibrium with its liquid phase at a specific temperature in a closed system.

To calculate solubility of a dissolves gas, Henry's Law Constant and partial pressure are used.

Solutions: Mass Percent

Mass Percent represents grams of solute per grams of solution, multiplied by 100.

Osmosis

Osmosis is the net movement of solvent across a semipermeable membrane.

Parts per Million (ppm)

<div dir="auto"><div contenteditable="true"dir="auto"spellcheck="true">Parts per Million (ppm) and Parts per Billion (ppb) are used to express extremely dilute solution concentrations.</div></div>