A solution is made by dissolving 21.5 grams of glucose (C6H12O6) in 255 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m?

nbklgyhlo

i wish i knew

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

ΔTf = Kf * m

Where:
ΔTf = freezing-point depression
Kf = freezing-point constant
m = molality of the solution

First, we need to calculate the molality of the solution.

Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is glucose (C6H12O6), and the solvent is water.

Step 1: Calculate the number of moles of glucose (C6H12O6)
To find the number of moles, we can divide the mass of glucose by its molar mass.
Molar mass of glucose (C6H12O6) = (6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol) = 180.18 g/mol

Number of moles of glucose = mass of glucose / molar mass
= 21.5 g / 180.18 g/mol

Step 2: Calculate the mass of water in kilograms.
To convert grams to kilograms, we divide by 1000:
mass of water = 255 g / 1000 = 0.255 kg

Step 3: Calculate the molality (m).
Molality (m) = moles of solute / mass of solvent (in kg)
= moles of solute / mass of water
= 21.5 g / (0.255 kg)

Now that we have the molality (m), we can calculate the freezing-point depression using the freezing-point constant (Kf) provided.

ΔTf = Kf * m
= -1.86 °C/m * (21.5 g / 0.255 kg)

Calculate the value to find the freezing-point depression (ΔTf).