বরফের গলনাঙ্ক হল......... কেলভিন। [273/373] | 9 | গ্যাস আইন অধ্যয়ন | রসায়ন ...

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 kg mol^(-1)) Freezing point depression constant to ethanol (K_(f)^(ethanol))=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)^(ethanol))=1.2 K 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 Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In anwering 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 mole fraction of water in the solution becomes 0.9 . The boiling point of this solution is

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 kg mol^(-1)) Freezing point depression constant to ethanol (K_(f)^(ethanol))=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)^(ethanol))=1.2 K 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 Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In anwering 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

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 kg mol^(-1)) Freezing point depression constant to ethanol (K_(f)^(ethanol))=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)^(ethanol))=1.2 K 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 Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In anwering the following questions consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The freezing point 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 :