Ethylene glycol (molar mass = 62 g/mol)is used as an antifreeze in cars. if 450 g of ethylene glycol is added to 4.25 kg of water, what is the molality? calculate how much the freezing point of water will be lowered. the freezing point depression constant for water is kf = -1.86°c/m. show your work.

Molality is 1.7m and the freezing point of water will be lowered by

Explanation:

Depression in freezing point is given by:

= Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

= freezing point constant =

m= molality  =

Weight of solvent (water) = 4.25 kg  

Molar mass of solute (ethylene glycol) = 62 g/mol

Mass of solute added (ethylene glycol) = 450 g

m= molality  =

Molality of solution will be 1.7m.

The freezing point of water will be lowered by

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%