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

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.