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What is the freezing point of a solution of 0.5 mol of LiBr in 500 mL of water (kf= 1.86 C/m)?

a. -3.72 C
b. -0.93 C
c. -1.86 C
d. -2.79 C"

1 Answer

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Final answer:

The freezing point of the solution is calculated using the formula for freezing point depression and the given molality. It is found to be -3.72°C, which is answer option (a).

Step-by-step explanation:

The question requires an understanding of the concept of freezing point depression in solutions, which is governed by the formula ΔTf = i * Kf * m, where i is the van 't Hoff factor (number of particles the solute splits into), Kf is the cryoscopic constant of the solvent (for water it is 1.86 °C/m), and m is the molality of the solution. Since LiBr splits into two ions, Li+ and Br-, its van 't Hoff factor (i) is 2.

The molality of the solution is calculated as moles of solute per kilogram of solvent. Here, the molality (m) is 0.5 mol / 0.5 kg = 1 mol/kg. Next, we use the formula above: ΔTf = 2 * 1.86 °C/m * 1 m = 3.72 °C.

Therefore, the freezing point of the water will decrease by 3.72 °C. Since the normal freezing point of water is 0.0 °C, the new freezing point will be 0.0 °C - 3.72 °C = -3.72 °C. Thus, the correct answer is (a) -3.72 °C.

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