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What is the freezing point, in C, of a 2.75 m solution of C8H18 in benzene?

What is the freezing point, in C, of a 2.75 m solution of C8H18 in benzene?-example-1
User Irving
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The freezing point of the 2.75 m solution of octane in benzene is -8.56 °C.

The freezing point of a solution is the temperature at which the solution becomes a solid. The freezing point of a solution is lower than the freezing point of the pure solvent because the solute particles interfere with the movement of the solvent molecules, which slows down the freezing process.

To determine the freezing point of a solution, we can use the freezing point depression equation:

ΔTf = Kf x molality

where ΔTf is the change in freezing point, Kf is the freezing point depression constant for the solvent, and molality is the concentration of the solute in the solution expressed in moles of solute per kilogram of solvent.

To find the freezing point of a 2.75 m solution of C8H18 (octane) in benzene, we need to know the freezing point depression constant for benzene, which is 5.12 °C/m. We can then use the equation above to calculate the change in freezing point:

ΔTf = 5.12 °C/m x 2.75 m = 14.06 °C

To find the freezing point of the solution, we need to subtract the change in freezing point from the freezing point of the pure solvent. The freezing point of pure benzene is 5.5 °C, so the freezing point of the 2.75 m solution of octane in benzene is:

5.5 °C - 14.06 °C = -8.56 °C

This means that the freezing point of the 2.75 m solution of octane in benzene is -8.56 °C. At this temperature, the solution will become a solid.

User JLK
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