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Paladinen [302]
4 years ago
8

What is the freezing point of a 0.743m aqueous solution if KCI? (Report amount to three decimal points)

Chemistry
1 answer:
Aleksandr-060686 [28]4 years ago
5 0
This is a question about the colligative property known as freezing point depression. Freezing point depression (the amount the normal freezing point of the solvent is decreased) can be calculated with this equation:

ΔT = i Kf<span> m
</span>
Where i (the van't Hoff factor) is the degree of dissociation of the solute, Kf is the freezing point depression constant, and m is the molality of the solution.

Here i = 2 (KCl dissociates into 2 ions, K+ and Cl-), Kf = 1.86 C/m (for water), and m = 0.743m).

ΔT = 2 x 1.86 C/m x 0.743m = <span>2.764C
</span>
That means the freezing point of the solution is 2.764C less than the pure solvent (water), making it 0C - 2.764C = -2.764C.
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3 years ago
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4 0
3 years ago
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tangare [24]
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3 0
3 years ago
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3 years ago
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Andrei [34K]

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Just like the question says, the Non-covalent bonds, ''makes it possible for a macromolecule to interact with great specificity with just one out of the many thousands of different molecules present inside a cell".

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7 0
3 years ago
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