What is the freezing point of a solution that contains 10.0 g of glucose (C₆H₁₂O₆)in 100.g of H₂O? K_f for water is 1.86°C/m. A)-0.186°C B)+0.186°C C)-0.10°C D)+0.10°C E)-1.03°C

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The freezing point depression can be calculated using the formula ΔTf = i * Kf * m, where i is the van't Hoff factor (1 for glucose), Kf is the freezing point depression constant (1.86°C/m for water), and m is the molality of the solution. First, calculate the molality: moles of glucose = 10.0 g / 180.18 g/mol = 0.0555 mol; molality = 0.0555 mol / 0.1 kg = 0.555 m. ΔTf = 1 * 1.86°C/m * 0.555 m = 1.03°C. The freezing point is lowered by 1.03°C, so the new freezing point is -1.03°C.

What Is the Freezing Point of a Solution That Contains