Asolution is made by dissolving 25.5 grams of glucose (c6h12o6) in 398 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.
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The freezing point depression is a coligative property.

The formula for the freezing point depression is:

ΔTf = Kf * m

Where m is the molality and Kf is the cryoscopic constant.

Kf of water is 1.86 °C / m

m = moles of solute / kg of solvent

moles of solute = mass in grams / molar mass

molar mass of C6H12O6 = 6*12g/mol + 12*1g/mol + 6*16g/mol = 180 g/mol

moles of C6H12O6 = 25.5 g / 180 g/mol = 0.14167 mol

m = 0.14167 mol / 0.398 kg = 0.3559 m

=> ΔT = 1.86°C/m * 0.3559m = 0.66°C

0.66°C