Asolution 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.

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

You use the atomic number to find out how many protons and neutrons there are in the atom. you use the atomic mass to find out how many protons and electrons there are in the atom.