$K_f$ for water is$1.86Kkg{mol}^{-1}$. If your automobile radiator holds 1.0 kg of water ,how many grams of ethylene glycol $\left(C_2{H}_6{O}_2\right)$ must you add to get the freezing point of the solution lowered to $-2.8^{o}C$?

Q. If a automobile radiator holds 1.0 Kg of water; how many grams of ethylene glycol $\left(C_2{H}_6{O}_2\right)$ must be add to get the freezing point of the solution lowered to $-{2.79}^{\circ }C$?
[$K_f$ for water is $1.86Kkgmo{l}^{-1}$]

Q. Value of kf of water is 1.86 k g mole ^-1 if your vehicles radiator can accommodate 1kg of water then how many grams of ethylene glycol can be dissolved in it so that freezing point of solution becomes -2.8 degree Celsius

Q. How many grams of $C_2{H}_6{O}_2$ (ethylene glycol) must be added to $1\text{kg}$ of water to produce a solution that freezes at –$3^{\circ }\text{C}.K_f\text{of water}={1.86}^{\circ }\text{C/molal}$.

Q. The freezing point of a solution containing $50c{m}^3$ of ethylene glycol in 50 g of water is found to be $-4^{o}C$. Assuming ideal behaviour, the density of ethylene glycol $(K_f$ for water $=1.86Kkgmo{l}^{-1}$) is:

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$-{24}^{\circ }\text{C}.(1\text{gallon}=3.785\text{lit},K_f=1.86)$?