Calculate the freezing point of a solution that contains 8.0 g of sucrose (c12h22o11) in 100 g of h2o. kf for h2o = 1.86c/m

Answer is: the freezing point of the solution of sucrose is -0.435°C.
m(H₂O) = 100 g ÷ 1000 g/kg = 0.1 kg.
m(C₁₂H₂₂O₁₁) = 8.0 g. 
n(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 8.0 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0233 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0233 mol ÷ 0.1 kg.
b(solution) = 0.233 m.
ΔT = b(solution) · Kf(H₂O).
ΔT = 0.233 m · 1.86°C/m.
ΔT = 0.435°C.
Tb = 0°C - 0.435°C = -0.435°C.

What do you mean by that? like the contents? the thc level?

Green is technically every other color but green but the other wavelengths of the other colors get absorbed by the object and the wavelength of the color green gets reflected