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Q1.12:
State Henry’s law and mention some important applications.
Solution :
Step 1: Definition and Theoretical Foundation of Henry's Law
Henry’s Law provides a quantitative relationship between the pressure of a gas above a liquid and the concentration of the gas dissolved in that liquid at a constant temperature. It states that: "The partial pressure of the gas in the vapour phase ($p$) is proportional to the mole fraction of the gas ($x$) in the solution."
Mathematically, this is expressed as:
$p = K_H \cdot x$
Where:
- $p$ = Partial pressure of the gas above the liquid solution.
- $x$ = Mole fraction of the gas in the solution (representing solubility).
- $K_H$ = Henry’s Law constant, which is a function of the nature of the gas and the temperature [Per the kinetic theory of gases and intermolecular force analysis].
Note: As $K_H$ increases at a given pressure, the solubility of the gas decreases. Furthermore, since the dissolution of gas in a liquid is an exothermic process, $K_H$ increases with rising temperature, explaining why aquatic species are more comfortable in cold waters.
Step 2: Graphical Representation of Henry's Law
Step 3: Important Applications of Henry's Law
| Application Area | Mechanism and Justification |
|---|---|
| Carbonated Beverages | Bottles of soft drinks and soda are sealed under high pressure to increase the solubility of $CO_2$ in the liquid [Application of the direct proportionality between $p$ and $x$]. |
| Scuba Diving | Divers face the "bends" (decompression sickness) due to high external pressure underwater, forcing atmospheric nitrogen to dissolve in blood. Upon ascent, decreasing pressure causes gas bubbles to form, blocking capillaries [Physiological consequence of shifting gas solubility]. |
| High Altitude Physiology | At higher altitudes, partial pressure of $O_2$ is lower than at ground level. This leads to low blood oxygen levels in climbers, causing hypoxia and an inability to think clearly [Per Henry's Law, lower $p$ results in lower dissolved $O_2$ in tissues]. |
Step 4: Summary of Theoretical Significance
Henry's Law is a specific case of Raoult's Law, where the solute is a gas. It serves as a fundamental principle in chemical engineering, environmental science, and human physiology. It dictates that for a given solvent-solute system at constant temperature, the equilibrium state is defined strictly by the pressure head applied to the gas phase.
Final Solution: Henry's Law states $p = K_H \cdot x$. Its critical applications include the pressurization of carbonated drinks, the management of blood gas concentration in scuba diving, and the physiological impact of low atmospheric oxygen at high altitudes.
More Questions from Class 12 Chemistry Solutions Exercises
- Q1.1: Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
- Q1.1: Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
- Q1.10: What role does the molecular interaction play in a solution of alcohol and water?
- Q1.10: What role does the molecular interaction play in a solution of alcohol and water?
- Q1.11: Why do gases always tend to be less soluble in liquids as the temperature is raised?
- Q1.11: Why do gases always tend to be less soluble in liquids as the temperature is raised?
- Q1.12: State Henry’s law and mention some important applications.
- Q1.13: The partial pressure of ethane over a solution containing \(6.56 \times 10^{-3}\) g of ethane is 1 bar. If the solution contains \(5.00 \times 10^{-2}\) g of ethane, then what shall be the partial pressure of the gas?
- Q1.13: The partial pressure of ethane over a solution containing (6.56 \times 10^{-3}) g of ethane is 1 bar. If the solution contains (5.00 \times 10^{-2}) g of ethane, then what shall be the partial pressure of the gas?
- Q1.14: What is meant by positive and negative deviations from Raoult's law and how is the sign of \(\Delta_{mix}H\) related to positive and negative deviations from Raoult's law?
- Q1.14: What is meant by positive and negative deviations from Raoult's law and how is the sign of (\Delta_{mix}H) related to positive and negative deviations from Raoult's law?
- Q1.15: An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
- Q1.15: An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
- Q1.16: Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
- Q1.16: Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
- Q1.17: The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
- Q1.17: The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
- Q1.18: Calculate the mass of a non-volatile solute (molar mass 40 g \(mol^{-1}\)) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
- Q1.18: Calculate the mass of a non-volatile solute (molar mass 40 (g mol^{-1})) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
- Q1.19 (i): A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute
- Q1.19 (ii): A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (ii) vapour pressure of water at 298 K.
- Q1.19(i): A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute
- Q1.19(ii): A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (ii) vapour pressure of water at 298 K.
- Q1.2: Give an example of a solid solution in which the solute is a gas.
- Q1.2: Give an example of a solid solution in which the solute is a gas.
- Q1.20: A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
- Q1.20: A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
- Q1.21: Two elements \(A\) and \(B\) form compounds having formula \(AB_{2}\) and \(AB_{4}\). When dissolved in 20 g of benzene (\(C_{6}H_{6}\)), 1 g of \(AB_{2}\) lowers the freezing point by 2.3 K whereas 1.0 g of \(AB_{4}\) lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg \(mol^{-1}\). Calculate atomic masses of \(A\) and \(B\).
- Q1.21: Two elements (A) and (B) form compounds having formula (AB_{2}) and (AB_{4}). When dissolved in 20 g of benzene ((C_{6}H_{6})), 1 g of (AB_{2}) lowers the freezing point by 2.3 K whereas 1.0 g of (AB_{4}) lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol–1. Calculate atomic masses of (A) and (B).
- Q1.22: At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
- Q1.22: At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
- Q1.23 (i): Suggest the most important type of intermolecular attractive interaction in the following pairs. (i) n-hexane and n-octane
- Q1.23 (ii): Suggest the most important type of intermolecular attractive interaction in the following pairs. (ii) \(I_{2}\) and \(CCl_{4}\)
- Q1.23 (iii): Suggest the most important type of intermolecular attractive interaction in the following pairs. (iii) \(NaClO_{4}\) and water
- Q1.23 (iv): Suggest the most important type of intermolecular attractive interaction in the following pairs. (iv) methanol and acetone
- Q1.23 (v): Suggest the most important type of intermolecular attractive interaction in the following pairs. (v) acetonitrile (\(CH_{3}CN\)) and acetone (\(C_{3}H_{6}O\)).
- Q1.23(i): Suggest the most important type of intermolecular attractive interaction in the following pairs. (i) n-hexane and n-octane
- Q1.23(ii): Suggest the most important type of intermolecular attractive interaction in the following pairs. (ii) (I_{2}) and (CCl_{4})
- Q1.23(iii): Suggest the most important type of intermolecular attractive interaction in the following pairs. (iii) (NaClO_{4}) and water
- Q1.23(iv): Suggest the most important type of intermolecular attractive interaction in the following pairs. (iv) methanol and acetone
- Q1.23(v): Suggest the most important type of intermolecular attractive interaction in the following pairs. (v) acetonitrile ((CH_{3}CN)) and acetone ((C_{3}H_{6}O)).
- Q1.24: Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, \(CH_{3}OH\), \(CH_{3}CN\).
- Q1.24: Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, (KCl), (CH_{3}OH), (CH_{3}CN).
- Q1.25 (i): Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i) phenol
- Q1.25 (ii): Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (ii) toluene
- Q1.25 (iii): Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (iii) formic acid
- Q1.25 (iv): Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (iv) ethylene glycol
- Q1.25 (v): Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (v) chloroform
- Q1.25 (vi): Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (vi) pentanol.
- Q1.25(i): Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i) phenol
- Q1.25(ii): Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (ii) toluene
- Q1.25(iii): Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (iii) formic acid
- Q1.25(iv): Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (iv) ethylene glycol
- Q1.25(v): Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (v) chloroform
- Q1.25(vi): Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (vi) pentanol.
- Q1.26: If the density of some lake water is 1.25g \(mL^{-1}\) and contains 92 g of \(Na^{+}\) ions per kg of water, calculate the molarity of \(Na^{+}\) ions in the lake.
- Q1.26: If the density of some lake water is 1.25g (mL^{-1}) and contains 92 g of (Na^{+}) ions per kg of water, calculate the molarity of (Na^{+}) ions in the lake.
- Q1.27: If the solubility product of CuS is \(6 \times 10^{-16}\), calculate the maximum molarity of CuS in aqueous solution.
- Q1.27: If the solubility product of CuS is (6 \times 10^{-16}), calculate the maximum molarity of CuS in aqueous solution.
- Q1.28: Calculate the mass percentage of aspirin (\(C_{9}H_{8}O_{4}\)) in acetonitrile (\(CH_{3}CN\)) when 6.5 g of \(C_{9}H_{8}O_{4}\) is dissolved in 450 g of \(CH_{3}CN\).
- Q1.28: Calculate the mass percentage of aspirin ((C_{9}H_{8}O_{4})) in acetonitrile ((CH_{3}CN)) when 6.5 g of (C_{9}H_{8}O_{4}) is dissolved in 450 g of (CH_{3}CN).
- Q1.29: Nalorphene (\(C_{19}H_{21}NO_{3}\)), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of \(1.5 \times 10^{-3}\) m aqueous solution required for the above dose.
- Q1.29: Nalorphene ((C_{19}H_{21}NO_{3})), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of (1.5 \times 10^{-3} m) aqueous solution required for the above dose.
- Q1.3 (i): Define the following terms: (i) Mole fraction
- Q1.3 (ii): Define the following terms: (ii) Molality
- Q1.3 (iii): Define the following terms: (iii) Molarity
- Q1.3 (iv): Define the following terms: (iv) Mass percentage.
- Q1.3(i): Define the following terms: (i) Mole fraction
- Q1.3(ii): Define the following terms: (ii) Molality
- Q1.3(iii): Define the following terms: (iii) Molarity
- Q1.3(iv): Define the following terms: (iv) Mass percentage.
- Q1.30: Calculate the amount of benzoic acid (\(C_{6}H_{5}COOH\)) required for preparing 250 mL of 0.15 M solution in methanol.
- Q1.30: Calculate the amount of benzoic acid ((C_{6}H_{5}COOH)) required for preparing 250 mL of 0.15 M solution in methanol.
- Q1.31: The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
- Q1.31: The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
- Q1.32: Calculate the depression in the freezing point of water when 10 g of \(CH_{3}CH_{2}CHClCOOH\) is added to 250 g of water. \(K_{a} = 1.4 \times 10^{-3}\), \(K_{f} = 1.86\) K kg \(mol^{-1}\).
- Q1.32: Calculate the depression in the freezing point of water when 10 g of (CH_{3}CH_{2}CHClCOOH) is added to 250 g of water. (K_{a} = 1.4 \times 10^{-3}), (K_{f} = 1.86 K kg mol^{-1}).
- Q1.33: 19.5 g of \(CH_{2}FCOOH\) is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
- Q1.33: 19.5 g of (CH_{2}FCOOH) is dissolved in 500 g of water. The depression in the freezing point of water observed is (1.0^{\circ}C). Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
- Q1.34: Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
- Q1.34: Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
- Q1.35: Henry’s law constant for the molality of methane in benzene at 298 K is \(4.27 \times 10^{5}\) mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
- Q1.35: Henry’s law constant for the molality of methane in benzene at 298 K is (4.27 \times 10^{5}) mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
- Q1.36: 100 g of liquid A (molar mass 140 g \(mol^{-1}\)) was dissolved in 1000 g of liquid B (molar mass 180 g \(mol^{-1}\)). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
- Q1.36: 100 g of liquid A (molar mass 140 (g mol^{-1})) was dissolved in 1000 g of liquid B (molar mass 180 (g mol^{-1})). The vapour pressure of pure liquid (B) was found to be 500 torr. Calculate the vapour pressure of pure liquid (A) and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
- Q1.37: Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot \(p_{total}\), \(p_{chloroform}\), and \(p_{acetone}\) as a function of \(x_{acetone}\). The experimental data observed for different compositions of mixture is: Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.
- Q1.37: Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot (p_{total}), (p_{chloroform}), and (p_{acetone}) as a function of (x_{acetone}). The experimental data observed for different compositions of mixture is: Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.
- Q1.38: Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.
- Q1.38: Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.
- Q1.39: The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are \(3.30 \times 10^{7}\) mm and \(6.51 \times 10^{7}\) mm respectively, calculate the composition of these gases in water.
- Q1.39: The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are (3.30 \times 10^{7}) mm and (6.51 \times 10^{7}) mm respectively, calculate the composition of these gases in water.
- Q1.4: Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g \(mL^{-1}\)?
- Q1.4: Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL–1?
- Q1.40: Determine the amount of \(CaCl_{2}\) (\(i\) = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27° C.
- Q1.40: Determine the amount of (CaCl_{2}) ((i = 2.47)) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at (27^{\circ} C).
- Q1.41: Determine the osmotic pressure of a solution prepared by dissolving 25 mg of \(K_{2}SO_{4}\) in 2 litre of water at 25° C, assuming that it is completely dissociated.
- Q1.41: Determine the osmotic pressure of a solution prepared by dissolving 25 mg of (K_{2}SO_{4}) in 2 litre of water at (25^{\circ} C), assuming that it is completely dissociated.
- Q1.5: A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g \(mL^{-1}\), then what shall be the molarity of the solution?
- Q1.5: A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL–1, then what shall be the molarity of the solution?
- Q1.6: How many mL of 0.1 M HCl are required to react completely with 1 g mixture of \(Na_{2}CO_{3}\) and \(NaHCO_{3}\) containing equimolar amounts of both?
- Q1.6: How many mL of 0.1 M HCl are required to react completely with 1 g mixture of (Na_{2}CO_{3}) and (NaHCO_{3}) containing equimolar amounts of both?
- Q1.7: A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
- Q1.7: A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
- Q1.8: An antifreeze solution is prepared from 222.6 g of ethylene glycol (\(C_{2}H_{6}O_{2}\)) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g \(mL^{-1}\), then what shall be the molarity of the solution?
- Q1.8: An antifreeze solution is prepared from 222.6 g of ethylene glycol ((C_{2}H_{6}O_{2})) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?
- Q1.9 (i): A sample of drinking water was found to be severely contaminated with chloroform (\(CHCl_{3}\)) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (i) express this in percent by mass
- Q1.9 (ii): A sample of drinking water was found to be severely contaminated with chloroform (\(CHCl_{3}\)) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (ii) determine the molality of chloroform in the water sample.
- Q1.9(i): A sample of drinking water was found to be severely contaminated with chloroform ((CHCl_{3})) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (i) express this in percent by mass
- Q1.9(ii): A sample of drinking water was found to be severely contaminated with chloroform ((CHCl_{3})) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (ii) determine the molality of chloroform in the water sample.
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