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Vitek1552 [10]
2 years ago
6

What is the freezing point of an aqueous 1.58 molal nacl solution?

Chemistry
1 answer:
Alina [70]2 years ago
4 0
<span>Tf is the freezing point of the solution(the solvent plus solute). T*f is the freezing point of the pure solvent(without solute) i is the van't Hoff factor.It is approximately the number of particles in solution that are made for each particle of the solute that is placed into solution.Therefore, for nonelectrolytes, i = 1. Kf is the freezing point depression constant.For water, Kf = 1.86 Degree C/m, or 1.86 Degree C.kg/mol. Tf is -1.58 Degree C</span>
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<em>∴ m = (mass/molar mass of solute)/(mass of water (kg)).</em>

<em />

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∵ m = (mass/molar mass of CaCl₂·6H₂O)/(mass of water (kg)).

m = 0.125 m, molar mass of CaCl₂·6H₂O = 219.0757 g/mol, mass of water = 500.0 g = 0.5 kg.

∴ 0.125 m = (mass of CaCl₂·6H₂O / 219.0757 g/mol)/(0.5 kg).

∴ mass of CaCl₂·6H₂O = (0.125 m)(219.0757 g/mol)(0.5 kg) = 13.7 g.

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∵ m = (mass/molar mass of NiSO₄·6H₂O)/(mass of water (kg)).

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