QuestionJuly 10, 2025

What is the freezing point of a solution of 0.5 mol of LiBr in 500 mL of water? (K^f=1.86^circ C/m) -7.44C -3.72C -5.58C -1.86C

What is the freezing point of a solution of 0.5 mol of LiBr in 500 mL of water? (K^f=1.86^circ C/m) -7.44C -3.72C -5.58C -1.86C
What is the freezing point of a solution of 0.5 mol of LiBr in 500 mL of water? (K^f=1.86^circ C/m) -7.44C -3.72C -5.58C -1.86C

Solution
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Answer

-3.72^\circ C Explanation 1. Calculate molality Molality m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{0.5 \, \text{mol}}{0.5 \, \text{kg}} = 1 \, m. 2. Determine van't Hoff factor LiBr dissociates into Li⁺ and Br⁻ ions, so the van't Hoff factor i = 2. 3. Apply freezing point depression formula **\Delta T_f = i \cdot K_f \cdot m**. Substituting values: \Delta T_f = 2 \cdot 1.86 \, ^\circ C/m \cdot 1 \, m = 3.72 \, ^\circ C. 4. Calculate new freezing point The normal freezing point of water is 0^\circ C. New freezing point = 0^\circ C - 3.72^\circ C = -3.72^\circ C.

Explanation

1. Calculate molality<br /> Molality $m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{0.5 \, \text{mol}}{0.5 \, \text{kg}} = 1 \, m$.<br /><br />2. Determine van't Hoff factor<br /> LiBr dissociates into Li⁺ and Br⁻ ions, so the van't Hoff factor $i = 2$.<br /><br />3. Apply freezing point depression formula<br /> **$\Delta T_f = i \cdot K_f \cdot m$**. Substituting values: $\Delta T_f = 2 \cdot 1.86 \, ^\circ C/m \cdot 1 \, m = 3.72 \, ^\circ C$.<br /><br />4. Calculate new freezing point<br /> The normal freezing point of water is $0^\circ C$. New freezing point $= 0^\circ C - 3.72^\circ C = -3.72^\circ C$.
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