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Lv2] Online 1 × e Calculate The Conce | + g.cengagenow.com/ilm/takeAssignment/ta M | Review Toplcs] [Re Use the References to access impertant values if needed for this question The freezing point of benzene, Cels, is 5.500 °C at 1 atmosphere. K(benzene)- 512°C/m In a laboratory experiment, students synthesized a new compound and found that when 14.44 grams of the compound were dissolved in 255.2 grams of benzene, the solution began to freeze at 4.495°C. The compound was also found to be nonvolatile and a non- electrolyte. What is the molecular weight they determined for this compound ? gmol Retry Entire Group g more group attempts remaining

Explanation / Answer

Answer 1:- Given data in the first question:

Freezing point of benzene (C6H6) = 5.500 0C (Also referred as solvent in calculation)

Kf of benzene = -5.12 0C/m

Freezing point of solution = 4.495 0C

Weight of new compound = 14.44 g.

Weight of benzene = 255.2 g = 0.2552 Kg.

Thus, the molecular weight of the new compound is calculated by getting the moles of solute using the concept of molality:

(5.500 0C - 4.495 0C) / (5.12 0C/m) = 0.19629 m

[ minus sign is not considered in the value of Kf as the F.P. of solution is lower than the F.P. of solvent ]

0.19629 m/Kg X 0.2552 Kg = 0.05009 m

Answer 2:- Again, in the second question, the freezing point of the solution is lower than that of the solvent ( ethanol, CH3CH2OH).

To calculate the molecular weight of the new compound, we will use the above concept:

{-117.300 0C - (-117.629 0C)} / (1.99 0C/m) = 0.16533 m

0.16533 m/Kg X 0.2822 Kg = 0.04665 m

Molecular weight of new compound = 14.44 g/0.05009 m = 288.28 g/mol.

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