Asolution of water (kf=1.86 ∘c/m) and glucose freezes at − 2.15 ∘c. what is the molal concentration of glucose in this solution? assume that the freezing point of pure water is 0.00 ∘c.

1.16 moles/liter The equation for freezing point depression in an ideal solution is ΔTF = KF * b * i where ΔTF = depression in freezing point, defined as TF (pure) â’ TF (solution). So in this case ΔTF = 2.15 KF = cryoscopic constant of the solvent (given as 1.86 âc/m) b = molality of solute i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1. Solving for b, we get ΔTF = KF * b * i ΔTF/KF = b * i ΔTF/(KF*i) = b And substuting known values. ΔTF/(KF*i) = b 2.15âc/(1.86âc/m * 1) = b 2.15/(1.86 1/m) = b 1.155913978 m = b So the molarity of the solution is 1.16 moles/liter to 3 significant figures.

Okay well gravitational energy is matter forced by gravity making energy though the kinetic energy then the electric energy produces electricity from the kinetic energy and the kinetic energy gets its energy from the gravitational energy. was that the answer you wanted or something else?