freezing point depression

Question

0.100 kg water contains 0.010 mol glucose (i=1). Kf(water)=1.86 K kg mol−¹. ΔTf is:

RankGuru · Accepted Answer

1. **Step 1: Calculate Molality ($m$)**:
   - $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}}$
   - $m = \frac{0.010 	ext{ mol}}{0.100 	ext{ kg}} = 0.10 	ext{ mol/kg}$.
2. **Step 2: Use Colligative Equation**:
   - $\Delta T_f = i 	imes K_f 	imes m$
   - For Glucose ($non-electrolyte$), $i = 1$.
3. **Calculation**:
   - $\Delta T_f = 1 	imes 1.86 	ext{ K kg mol}^{-1} 	imes 0.10 	ext{ mol kg}^{-1}$
   - $\Delta T_f = 0.186 	ext{ K}$.
4. **Note**: The freezing point would be $0^\circ C - 0.186 K = -0.186^\circ C$.

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