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For a 0.222 m aqueous solution of sucrose, C₁₂H₂₂O₁₁, What is the boiling point?

kb = 0.512˚C/m

User Amena
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1 Answer

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Final answer:

The boiling point of a 0.222 m aqueous solution of sucrose can be calculated using the formula ΔTb = kb * m, where ΔTb is the boiling point elevation, kb is the molal boiling point elevation constant for water, and m is the molality of the solution. By substituting the given values into the formula, we find that the boiling point elevation is 0.113664 ˚C. Adding this to the boiling point of water, we get a new boiling point of 100.113664 ˚C.

Step-by-step explanation:

The boiling point of a solution can be calculated using the formula:



ΔTb = kb * m



Where ΔTb is the boiling point elevation, kb is the molal boiling point elevation constant for water, and m is the molality of the solution.



In this case, we are given a 0.222 m aqueous solution of sucrose. By substituting the values into the formula, we can calculate the boiling point elevation:



ΔTb = 0.512˚C/m * 0.222 m = 0.113664 ˚C



To find the boiling point of the solution, we add the boiling point elevation to the boiling point of water:



New boiling point = 100.00 ˚C + 0.113664 ˚C = 100.113664 ˚C

User James Blake
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