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A 1.35 m aqueous solution of compound X had a boiling point of 102.8°C. Which one of the following could be compound X? The boiling point elevation constant for water is 0.52°C/m.

a) C₆H12O₆
b) CaCl₂
c) CH₃CH₂OH
d) Na₃PO₄

User TheWolf
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1 Answer

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Final answer:

The boiling point elevation constant for water is 0.52°C/m. When a solute such as glucose or sucrose is added to water, the boiling point increases by 0.52°C for every 1 mole of solute per liter of water. However, when a solute like NaCl is added, which produces 2 moles of dissolved ions per mole of solute, the boiling point increase is approximately twice as large, the correct option is D).

Step-by-step explanation:

The boiling point of a solution is determined by the boiling point elevation constant and the molality of the solute. The boiling point elevation constant for water is 0.52°C/m. When a nonvolatile molecular solute such as glucose or sucrose is added to water, the boiling point increases by 0.52°C for every 1 mole of solute per liter of water.

However, when a solute like NaCl is added, which produces 2 moles of dissolved ions per mole of solute, the boiling point increase is approximately twice as large. Therefore, option d) Na₃PO₄ could be the compound X since it produces 3 moles of dissolved ions per mole of solute.

User Nandita Sharma
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