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Otrada [13]
3 years ago
9

A chemist dissolves 1.3 mol NaCl in a 2.0-kg sample of water. What is the boiling point of this solution? (Assume that pure wate

r boils at 100°C and that Kb for water = 0.512°C/m.)
Chemistry
2 answers:
mafiozo [28]3 years ago
8 0
For this problem, the solution is exhibiting some colligative properties since the solute in the solution interferes with some of the properties of the solvent. We use equation for the boiling point elevation for this problem. We do as follows:
<span>
ΔT(boiling point)  = (Kb)mi
</span>ΔT(boiling point)  = (0.512)(1.3/2.0)(2)
ΔT(boiling point)  = 0.67 degrees Celsius
<span>
T(boiling point) = 100 + 0.67 = 100.67 degrees Celsius</span>
lara31 [8.8K]3 years ago
7 0

Answer:

100.67°C

Explanation:

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Explanation:

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In this case, since no options are given we can infer from the statement that due to water's higher boiling point than acetone we can conclude that when they are in liquid state, water has stronger intermolecular forces which allow its particles to be held in a stronger way in comparison to the acetone's molecules, for that reason, more energy will be required in order to separate them and promote the boiling process, which is attained via increasing the temperature. Besides, less energy will be required for the separation of the acetone's molecules in order to boil it when liquid, therefore, a lower temperature is required.

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