Question

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.

Answer 1

Molality= moles solute/kg solvent

450g x 1mol/62 g = 7.26 moles

Molality = 7.26/4.24 kg = 1.71 m

Delta T = imK= (1) (1.71) (1.86)= 3 degrees

So the answer is that the freezing point should be lowered by 3 degrees

Hope this helps :)

Answer 2

Answer: Molality is 1.7m and the freezing point of water will be lowered by
`3.2^0C`

Step-by-step explanation:

Depression in freezing point is given by:

`\Delta T_f=i* K_f* m`

`\Delta T_f=T_f^0-T_f=(0-T_f)^0C` = Depression in freezing point

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

`K_f` = freezing point constant =
`1.86^0C/m`

m= molality =
`\frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}`

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 =
`(450g)/(62g/mol* 4.25kg)=1.7`

Molality of solution will be 1.7m.

`\Delta T_f=1* 1.86* 1.7`

`\Delta T_f=3.2^0C`

The freezing point of water will be lowered by
`3.2^0C`