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A solution of ethanol in water is 1.54 m...

A solution of ethanol in water is `1.54` molal. How many gams of ethanol are there in 2500g of the solvent ?

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To solve the problem of finding the grams of ethanol in a 1.54 molal solution with 2500 grams of water as the solvent, we can follow these steps: ### Step 1: Understand the definition of molality Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula for molality is: \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \] ### Step 2: Convert the mass of the solvent to kilograms Since the molality is given per kilogram of solvent, we need to convert the mass of the solvent (water) from grams to kilograms: ...
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Properties 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 ethanol and water. Thus moel 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 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 = 155.7K 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 Molecualr weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In asweering 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 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 ethanol and water. Thus moel 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 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 = 155.7K 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 Molecualr weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In asweering 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: