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Benzophenone has a normal freezing point of +48.1 oC, with freezing point depression constant Kfpt = − 9.78 oC/m. A 0.1500 molal solution of ionizing salt had a freezing point of +44.0 oC. What is the van't Hoff (ion dissociation) constant i for this salt? (i.e., the average number of ions produced in the solution.) Report 3 significant digits.

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anfabba15

Answer:

i = 2.79

Explanation:

The excersise talks about the colligative property, freezing point depression.

Formula to calculate the freezing point of a solution is:

Freezing point of pure solvent - Freezing point of solution = m . Kf . i

Let's replace data given. (i = Van't Hoff factor, numbers of ions dissolved in solution)

48.1°C - 44°C = 0.15 m . 9.78°C/m . i

4.1°C / (0.15 m . 9.78°C/m) = i

i = 2.79

In this case, numbers of ions dissolved can decrease the freezing point of a solution, which is always lower than pure solvent.

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