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An LPG cylinder weighs 14.8 kg when empt...

An `LPG` cylinder weighs `14.8 kg` when empty. When full it weighs `29.0 kg` and the weight of the full cylinder reduces to `23.2 kg`. Find out the volume of the gas in cubic metres used up at the normal usage conditions Assume `LPG` to be `n`-butane with normal boiling point of `0^(@)C`.

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To solve the problem step-by-step, we will follow these calculations: ### Step 1: Calculate the mass of gas consumed The mass of the gas consumed can be calculated by subtracting the weight of the full cylinder after usage from the weight of the full cylinder. \[ \text{Mass of gas consumed} = \text{Weight of full cylinder} - \text{Weight after usage} \] ...
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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. The vapour pressure 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

CENGAGE CHEMISTRY ENGLISH-STATES OF MATTER-Archives (Subjective)
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