A solution of two liquids, A and B, Shows negative deviation from Raoult\'s law.

Question

A solution of two liquids, A and B, Shows negative deviation from Raoult's law. This means that the molecules of A interact strongly with other A-type molecules the two liquids have a positive heat of solution. molecules of A interact weakly, if at all. with B molecules. the molecules of A hinder the strong interaction between B molecules. molecules of A interact more strongly with B than A with A or B with B. A 5.50-gram sample of a compound as dissolved in 250. grams of benzene. The freezing point of this solution is 1.02 degree C below that of pure benzene. What is the molar mass of this compound? 22.0 g/mol 110. g/mol 220. g/mol 44.0 g/mol none of these Thyroxine, an important hormone that controls the rate of metabolism in the body, can be isolated from the thyroid gland If 0.455 g of thyroxine is dissolved in 10.0 g of benzene the freezing point of the solution it 5.144 degree C Pure benzene freezes at 5 444 degree C and has a value for the molal freezing point depression constant of K_f, of 5.12 degree C/m. What is the molar mass of thyroxine? 777.000 g/mol 777 g/mol 2330 g/mol 285 g/mol 3760 g/mol To calculate the freezing point of an ideal dilute solution of a single, nondissociating solute of a solvent, the minimum information one must know is: the molality (of the solute) the molality (of the solute) and the freezing point depression constant of the solvent. the molality (of the solute), the freezing point depression constant of the solvent, and the freezing point of the pure solvent. the molality (of the solute), the freezing point depression constant of the solvent, the freezing point of the pure solvent, and the molecular weight of the solute. the molality (of the solute), the freezing point depression constant of the solvent, the freezing point of the pure solvent, and the weight of the solvent.

Explanation / Answer

Question 24. Answer-(E)

Negative deviations from Raoult's Law

In mixtures showing a negative deviation from Raoult's Law, the vapour pressure of the mixture is always lower than expect from an ideal mixture.

Explaining the deviations

The fact that the vapour pressure is lower than ideal in these mixtures means that molecules are breaking away less easily than they do in the pure liquids.

That is because the intermolecular forces between molecules of A and B are more than they are in the pure liquids.

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