Question

What is the freezing point depression of a solvent when 15.5 grams of glucose (C₆H₁₂O₆) are dissolved in 245 grams of water, considering the freezing point constant as -1.86 °C/m?

a) 0.1 °C
b) 1.0 °C
c) 2.5 °C
d) 3.5 °C
e) 5.0 °C

Answer 1

The freezing point depression of the solvent in the given solution is 0.65 °C.

Step-by-step explanation:

Freezing point depression is a colligative property that occurs when a solute is dissolved in a solvent. The freezing point depression is given by the equation ΔTf = Kf imes m, where ΔTf is the change in freezing point, Kf is the molal freezing point depression constant, and m is the molality of the solute. In this case, we can calculate the molality of the glucose solution by using the equation:

m = moles of solute / kg of solvent

Using the molar mass of glucose (180 g/mol), we can calculate the moles of glucose in the solution:

moles of glucose = 15.5 g / 180 g/mol = 0.0861 mol

Then, we can calculate the molality:

m = 0.0861 mol / 0.245 kg = 0.351 mol/kg

Substituting the values into the freezing point depression equation:

ΔTf = (-1.86 °C/m) imes (0.351 mol/kg) = -0.65 °C

Therefore, the freezing point depression of the solvent is 0.65 °C.