Question

A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 0.325 kilograms of water. If the molal freezing point constant for water (Kf) is -1.86 °C/m, what is the resulting Δtf of the solution? Show all of the steps taken to solve this problem.

Answer 1

I couldn't personally come up with the answer but I found a link the could help you! This is what I go to when I am having trouble finding an answer. 

https://www.scribd.com/document/358060502/12-Chem-1goalias-Blogspot-Com-Goalias-Blogspot1

Answer 2

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