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Properties such as boiling point, freezi...

Properties such as boiling point, freezing point, and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties.
Anwer the following questions:
i.`0.001 m NaCl`
ii.0.001 m urea
iii.`0.001 m MgCl_(2)`
iv.`0.001 m CH_(3)COOH`
Increasing order of boiling points

A

`(ii) lt (iv) lt (i) lt (iii)`

B

`(iv) lt (i) lt (ii) lt (iii)`

C

`(iii) lt (ii) lt (i) lt (iiv)`

D

`(i) lt (ii) lt (iii) lt (iv)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the increasing order of boiling points for the given solutions, we need to consider the van 't Hoff factor (i) for each solute, which represents the number of particles the solute dissociates into when dissolved in a solvent. The boiling point elevation (ΔTb) is directly proportional to the product of the van 't Hoff factor (i) and the molality (m) of the solution. ### Step-by-Step Solution: 1. **Identify the van 't Hoff factor (i) for each solute:** - For **NaCl**: It dissociates into 2 ions (Na⁺ and Cl⁻), so i = 2. - For **Urea**: It does not dissociate in solution, so i = 1. - For **MgCl₂**: It dissociates into 3 ions (Mg²⁺ and 2 Cl⁻), so i = 3. - For **Acetic Acid (CH₃COOH)**: It partially dissociates, but for simplicity, we can approximate i to be around 1.5 (since it is a weak acid and does not fully dissociate). 2. **Calculate the effective molality (i * m) for each solution:** - For **0.001 m NaCl**: \[ i \cdot m = 2 \cdot 0.001 = 0.002 \] - For **0.001 m Urea**: \[ i \cdot m = 1 \cdot 0.001 = 0.001 \] - For **0.001 m MgCl₂**: \[ i \cdot m = 3 \cdot 0.001 = 0.003 \] - For **0.001 m CH₃COOH**: \[ i \cdot m \approx 1.5 \cdot 0.001 = 0.0015 \] 3. **Rank the solutions based on the effective molality (i * m):** - Urea: 0.001 - Acetic Acid: 0.0015 - NaCl: 0.002 - MgCl₂: 0.003 4. **Determine the increasing order of boiling points:** The boiling point elevation is greater for solutions with higher effective molality. Therefore, the increasing order of boiling points is: \[ \text{Urea} < \text{Acetic Acid} < \text{NaCl} < \text{MgCl}_2 \] ### Final Answer: The increasing order of boiling points is: **0.001 m Urea < 0.001 m CH₃COOH < 0.001 m NaCl < 0.001 m MgCl₂**

To determine the increasing order of boiling points for the given solutions, we need to consider the van 't Hoff factor (i) for each solute, which represents the number of particles the solute dissociates into when dissolved in a solvent. The boiling point elevation (ΔTb) is directly proportional to the product of the van 't Hoff factor (i) and the molality (m) of the solution. ### Step-by-Step Solution: 1. **Identify the van 't Hoff factor (i) for each solute:** - For **NaCl**: It dissociates into 2 ions (Na⁺ and Cl⁻), so i = 2. - For **Urea**: It does not dissociate in solution, so i = 1. - For **MgCl₂**: It dissociates into 3 ions (Mg²⁺ and 2 Cl⁻), so i = 3. ...
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Properties such as boiling point, freezing point and vapour, pressure of a pure solvent change Propeties such as boiling point, freezing point and vapour, pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing athanol and water. The mole fraction of ethanol in the mixture is 0.9 Given Freezing point depression constant of water (K_(f)^("water"))=1.86 K kg "mol"^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol"))=2.0 K kg "mol"^(-1) Boiling point elevation constant of water (K_(b)^("water"))=0.52 kg "mol"^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethanol"))=1.2 kg "mol"^(-1) Standard freezing point of water =273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water =373 K tandard boiling point of ethanol =351.5 K Vapour pressure of pure water =32.8 mmHg Vapour presure of pure ethanol =40g Hg Molecular weight of water =18 g"mol"^(-1) Molecules weight of ethanol =46 g "mol"^(-1) In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing athanol and water. The mole fraction of ethanol in the mixture is 0.9 The freezing point of the solution M is

Properties such as boiling point, freezing point and vapour, pressure of a pure solvent change Propeties such as boiling point, freezing point and vapour, pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing athanol and water. The mole fraction of ethanol in the mixture is 0.9 Given Freezing point depression constant of water (K_(f)^("water"))=1.86 K kg "mol"^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol"))=2.0 K kg "mol"^(-1) Boiling point elevation constant of water (K_(b)^("water"))=0.52 kg "mol"^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethanol"))=1.2 kg "mol"^(-1) Standard freezing point of water =273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water =373 K tandard boiling point of ethanol =351.5 K Vapour pressure of pure water =32.8 mmHg Vapour presure of pure ethanol =40g Hg Molecular weight of water =18 g"mol"^(-1) Molecules weight of ethanol =46 g "mol"^(-1) In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing athanol and water. The mole fraction of ethanol in the mixture is 0.9 Given Freezing point depression constant of water (K_(f)^("water"))=1.86 K kg "mol"^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol"))=2.0 K kg "mol"^(-1) Boiling point elevation constant of water (K_(b)^("water"))=0.52 kg "mol"^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethanol"))=1.2 kg "mol"^(-1) Standard freezing point of water =273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water =373 K tandard boiling point of ethanol =351.5 K Vapour pressure of pure water =32.8 mmHg Vapour presure of pure ethanol =40g Hg Molecular weight of water =18 g"mol"^(-1) Molecules weight of ethanol =46 g "mol"^(-1) In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. Water is added to the solution M such that the fraction of water in the solution becomes 0.9 . The boiling point of this solutions is

Properties such as boiling point, freezing point and vapour, pressure of a pure solvent change Propeties such as boiling point, freezing point and vapour, pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing athanol and water. The mole fraction of ethanol in the mixture is 0.9 Given Freezing point depression constant of water (K_(f)^("water"))=1.86 K kg "mol"^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol"))=2.0 K kg "mol"^(-1) Boiling point elevation constant of water (K_(b)^("water"))=0.52 kg "mol"^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethanol"))=1.2 kg "mol"^(-1) Standard freezing point of water =273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water =373 K tandard boiling point of ethanol =351.5 K Vapour pressure of pure water =32.8 mmHg Vapour presure of pure ethanol =40g Hg Molecular weight of water =18 g"mol"^(-1) Molecules weight of ethanol =46 g "mol"^(-1) In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing athanol and water. The mole fraction of ethanol in the mixture is 0.9 Given Freezing point depression constant of water (K_(f)^("water"))=1.86 K kg "mol"^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol"))=2.0 K kg "mol"^(-1) Boiling point elevation constant of water (K_(b)^("water"))=0.52 kg "mol"^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethanol"))=1.2 kg "mol"^(-1) Standard freezing point of water =273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water =373 K tandard boiling point of ethanol =351.5 K Vapour pressure of pure water =32.8 mmHg Vapour presure of pure ethanol =40g Hg Molecular weight of water =18 g"mol"^(-1) Molecules weight of ethanol =46 g "mol"^(-1) In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The vapour pressure of the solution M is

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 . Given : Freezing point depression constant of water (K_(f)^("water")) = 1.86 K "mol"^(-1) Freezing point depression constant of ethanol (K_(f)^("ethonal")) = 2.0 K kg "mol"^(-1) Boiling point elevation constant of water (K_(b)^("water")) = 0.52 K kg "mol"^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethonal")) = 1.2 K kg mol^(-1) Standard freezing point of water = 273 K Standard freezing point of ethonal = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethonal = 40 mm Hg Molecular weight of water = 18 g "mol"^(-1) Molecular weight of ethonal = 45 g"mol"^(-1) In answering the following questions, consider the solution to be ideal ideal solutions and solutes to be non-volatile and non-dissociative. Water is added to the solution M such lthat the molecules fraction of water in t he solution becomes 0.9 . The boiling point of this solution is :

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 . Given : Freezing point depression constant of water (K_(f)^("water")) = 1.86 K "mol"^(-1) Freezing point depression constant of ethanol (K_(f)^("ethonal")) = 2.0 K kg "mol"^(-1) Boiling point elevation constant of water (K_(b)^("water")) = 0.52 K kg "mol"^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethonal")) = 1.2 K kg mol^(-1) Standard freezing point of water = 273 K Standard freezing point of ethonal = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethonal = 40 mm Hg Molecular weight of water = 18 g "mol"^(-1) Molecular weight of ethonal = 45 g"mol"^(-1) In answering the following questions, consider the solution to be ideal ideal solutions and solutes to be non-volatile and non-dissociative. Water is added to the solution M such lthat the molecules fraction of water in t he solution becomes 0.9 . The boiling point of this solution is :

Arrange the following solutions as: Increasing order of boiling points- 0.001 m NaCl , 0.001 m urea 0.001 m MgCl_(2) , 0.001 m CH_(3)COOH

For 0.1M solution ,the colligative property will follow the order

CENGAGE CHEMISTRY ENGLISH-SOLUTIONS-Exercises (Linked Comprehension)
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  8. A certain vessel X has water and nitrogen gas at a total pressure of 2...

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  9. A certain vessel X has water and nitrogen gas at a total pressure of 2...

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  10. A certain vessel X has water and nitrogen gas at a total pressure of 2...

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  11. A certain vessel X has water and nitrogen gas at a total pressure of 2...

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  12. A system of greater disorder of molecules is more probable. The disord...

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  14. A system of greater disorder of molecules is more probable. The disord...

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