3.35%(w/w)solution of ethylene glycol in water, an antifreezer used in automobiles in radiators as a coolant. It lowers freezing point of water to -17.6C .Calculate the mole fraction of the components.

Q. Ethylene glycol, $C{H}_{2}OH\cdot C{H}_{2}OH$, the major component of permanent antifreeze, effectively depresses the freezing point of water in an automobile radiator. What minimum weight of ethylene glycol must be mixed in kg with 6 gallons of water to protect it from freezing at$-{24}^{\circ }\text{C}.(1\text{gallon}=3.785\text{lit},K_f=1.86)$?

Q. In a cold climate, water gets frozen causing damage to the radiators of a car. Ethylene glycol is used as an antifreeze agent. The mass of ethylene glycol to be added to 4 kg water to prevent it from freezing at $-6^{o}C.K_f$ for water is $1.85Kmo{l}^{-1}kg$:

Q. In cold climate, water freezes causing damage to the radiator of the car . Ethylene glycol is used as an antifreeze. The amount of ethylene glycol to be added to 4 kg of water to prevent it from freezing at $-6^{o}C$ ($K_f$ of water = $1.85K{m}^{-1}$)

Q. Ethylene glycol is used as coolant in car radiators, in order to prevent the solution from freezing at -0.3${}^o$C. The amount of ethylene glycol to be added to 5 kg of water is (for water, $K_f=1.86K{m}^{-1})$:

Q. Ethylene glycol is used as an antifreeze agent. Calculate the amount of ethylene glycol to be added to $4kg$ of water to prevent it from freezing at $-6^{\circ }C.(K_f$ for $H_{2}O=1.85Kmol{e}^{-1}kg)$