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Molality of ethylene glycol, C₂H₄(OH)₂, in a solution prepared from 2. 331×10³ g of ethylene glycol and 2.00×10³ g of water, H₂O is 47.6m
Ethylene Glycol is known as C₂H₄(OH)₂. It is added in water to prepare an Antifreeze solution.
Given,
Mass of Ethylene Glycol = 2.331 × 10³ g = 2.331kg
Mass of Water = 2.00 × 10³ g
Since, Ethylene Glycol is in excess. Hence, it acts as a solvent and water acts as a solute.
We know, Molar Mass of Water = 18g
Hence, Moles of Water = Given mass of water / Molar Mass of Water
⇒ Moles of Water = 2000 / 18
⇒ Moles of Water = 111.1
Molality is defined as the moles of solute present in a given solvent in kg.
∴ Molality = Moles of Solute / Mass of Solvent (in kg)
Molality = Moles of Water / Mass of Ethylene Glycol
⇒ Molality = 111.1 / 2.331
⇒ Molality = 47.6m
Molality of ethylene glycol, C₂H₄(OH)₂, in a solution prepared from 2. 331×10³ g of ethylene glycol and 2.00×10³ g of water, H₂O is 47.6m
Learn more about Molality here, https://brainly.com/question/17034043
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