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Alina [70]
2 years ago
14

What is the 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?
Chemistry
1 answer:
qaws [65]2 years ago
3 0

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, brainly.com/question/17034043

#SPJ4

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