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.

To solve:
1.) Convert grams of glucose to moles of glucose:25.5g gluc. * (1 mol/180g gluc.) = 1.41E-1 mol gluc.      Note. Molar mass gluc. = 180g/mol
2.) Convert grams of solvent to kilograms of solvent:398 g H2O * (1kg H2O/1000g H2O) = 0.398 kg H2O
3.) Determine molality (moles solute/kilogram solvent) of solux:1.41E-1 mol gluc./0.398 kg H2O = 3.5427E-1 mol/kg
4.) Employ Freezing Point Depression equation and determine the freezing point of the solux:
ΔT = i Kf mΔT= ? (°C)Kf = 1.86 (°C kg mol¯1)m = 3.5427E-1 (mol/kg)i = (assumed negligible)
ΔΤ = (1.86)(3.5427E-1)ΔT = 6.589Ε-1°C
5.) ΔT = 6.589Ε-1°C
25.5g gluc. * (1 mol/180g gluc.) = equal 0.14166, not 1.41E-1 mol glucose

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