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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.
Solution :
Step 1: Theoretical Foundation of Freezing Point Depression
The depression in freezing point ($\Delta T_f$) is a colligative property defined by the relation:
$\Delta T_f = i \cdot K_f \cdot m$
[Per Raoult’s Law and the colligative properties of dilute solutions]. Here, $i$ represents the van't Hoff factor, $K_f$ is the cryoscopic constant of the solvent, and $m$ is the molality of the solution. Since the concentration ($m$) and the solvent ($K_f$) are constant for the three acids, the magnitude of $\Delta T_f$ is directly proportional to the van't Hoff factor ($i$).
Step 2: Analysis of the van't Hoff Factor ($i$)
For weak electrolytes (like organic acids), the van't Hoff factor is expressed as:
$i = 1 + \alpha(n - 1)$
Where $\alpha$ is the degree of dissociation and $n$ is the number of ions produced per molecule ($n=2$ for these monoprotic acids). Thus, $i = 1 + \alpha$. As the acid strength (tendency to dissociate into ions) increases, $\alpha$ increases, which consequently increases $i$ and $\Delta T_f$.
Step 3: Inductive Effect and Acidity Comparison
The acidity of carboxylic acids is determined by the stability of the conjugate base (carboxylate ion) formed after the release of a proton ($H^+$). Electron-withdrawing groups (EWGs) stabilize the negative charge on the carboxylate group via the inductive effect ($-I$ effect).
| Acid | Substituent | Inductive Effect |
|---|---|---|
| Acetic Acid ($CH_3COOH$) | $-CH_3$ | Weak $+I$ (Electron donating) |
| Trichloroacetic Acid ($CCl_3COOH$) | $-Cl$ (3 atoms) | Strong $-I$ |
| Trifluoroacetic Acid ($CF_3COOH$) | $-F$ (3 atoms) | Very Strong $-I$ |
Step 4: Logical Derivation of Dissociation Order
- Acetic Acid: The methyl group is electron-donating ($+I$ effect), which destabilizes the carboxylate ion, leading to minimal dissociation ($\alpha$ is smallest).
- Trichloroacetic Acid: Three chlorine atoms exert a strong $-I$ effect, stabilizing the conjugate base significantly, leading to higher dissociation than acetic acid.
- Trifluoroacetic Acid: Fluorine is more electronegative than chlorine. The three fluorine atoms exert the strongest $-I$ effect among the trio, providing the greatest stabilization to the conjugate base. Thus, $\alpha$ is largest.
Step 5: Visual Representation of Freezing Point Depression Mechanism
The dissociation constants ($K_a$) follow the order: $CH_3COOH < CCl_3COOH < CF_3COOH$.
Final Solution: The depression in freezing point increases in the order of acetic acid < trichloroacetic acid < trifluoroacetic acid because the electron-withdrawing inductive effect of the substituent groups increases in that order. This leads to an increase in the degree of dissociation ($\alpha$), a higher van't Hoff factor ($i$), and consequently a larger value for the colligative property $\Delta T_f$.
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.
- Q1.12: 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.
- Q1.32: 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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