F2 is light yellow/colorless in hydrocarbon solvent. i2 is pink. if a student combines fluorine water with nai (sodium iodide) in water, then adds pentane and observes a nearly colorless pentane layer, what can the student conclude? f is reduced from f0 to f- f is oxidized i- is oxidized to i0 i is reduced

F₂ + 2 NaI → 2 NaF + I₂
It is given that F₂ is light yellow / colorless in hydrocarbon solvent. The student combines Fluorine water with NaI in water. Then student adds pentane in the mixture of F₂ and NaI. After dissolution, solution was observed and a colorless pentane layer was seen. Alkanes are unreactive in nature. The C-H bond in alkane is difficult to break. whereas, F₂ is very reactive and reacts vigorously with alkanes in presence of light by free radical mechanism.It is given that the color of the solution is nearly colorless. F₂ when present in hydrocarbon solvent is light yellow/ colorless/ nearly colorless. Hence, F₂ is not reacting with hydrocarbon and there is no reaction taking place (No F₂ is present)

The answer to your question would be c. mineral oil  

Answer charle's law