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butalik
3 months ago
15

Determine the freezing point of a solution which contains 0.31 mol of sucrose in 175 g of water. kf = 1.86ï°c/m

Chemistry
1 answer:
eduard [2.7K]3 months ago
8 0
Freezing point depression is classified as a colligative property.

The governing formula is:

ΔT f = Kf * m

Here, m represents the molality of the solution => m = moles of solute / Kg of solvent

=> m = 0.31 mol / 0.175 kg = 1.771 m

And kf = 1.86 °C/m

=> ΔTf = 1.771 m * 1.86 °C / m = 3.3 °C

 => The solution's freezing point = normal freezing point of water - 3.3 °C = 0 - 3.3°C = - 3.3°C.

Answer: -3.3 °C


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Density is defined as the "mass per unit volume" of an object.

Thus, for an object weighing 100 grams with a volume of 100 milliliters, the density calculates to 100 grams / 100 ml.

When weighing the object, if there is water on the scale's surface, it will contribute additional weight, making the object seem heavier than its actual mass. Consequently, you might mistakenly conclude that the density is GREATER than it truly is.

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I hope this is helpful!

Best of luck
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