A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

To find the freezing-point depression of the solvent, you need to use the formula:

ΔT = K_f * m

where:
ΔT is the freezing-point depression,
K_f is the freezing point constant, and
m is the molality of the solution.

To find the molarity of the glucose solution:
1. Calculate the number of moles of glucose:

Molar mass of glucose (C6H12O6) = 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

Number of moles of glucose = mass of glucose / molar mass of glucose
= 15.5 g / 180.18 g/mol
= 0.086 moles

2. Calculate the mass of the solvent (water) in kilograms:

Mass of water = 245 g = 0.245 kg

3. Calculate the molality of the solution:

Molality (m) = moles of solute / mass of solvent (kg)
= 0.086 mol / 0.245 kg
= 0.351 mol/kg

4. Now, substitute the values into the freezing-point depression formula:

ΔT = K_f * m
= -1.86 °C/m * 0.351 mol/kg
= -0.65 °C

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