Question

This question has multiple parts. Work all the parts to get the most points. An aqueous antifreeze solution is 39.0% ethylene glycol () by mass. The density of the solution is 1.05 . a Calculate the molality of the ethylene glycol. Molality = mol/kg b Calculate the molarity of the ethylene glycol. Molarity = mol/L c Calculate the mole fraction of the ethylene glycol. Mole fraction =

Answer 1

Answer:

Molality = 10.300 m

Molarity = 6.5970 M

mole fraction = 0.156549

Explanation:

39.0 % = ethylene glycol

61.0 % = water

imagine the total mass = 100g

39.0% ethylene glycol = 39g

61.0 % water = 61g

1) Molality = number of moles / mass of solvent (kg)

Molar mass of ethylene glycol = 62.07g/mole

mole = 39g / 62.07g/mole = 0.6283 moles

Molality = moles / mass of solvent = 0.6283 moles / 0.061kg = 10.300 m

2) Molarity = number of moles / volume of solution

Since we know the density  of the solution = 1.05g /ml

⇒ volume = 100g / 1.05g /mL   = 95.24 mL = 0.09524 L

Molarity = 0.6283 moles / 0.09524 L = 6.5970 M

3) Mole fraction

moles water = 61g / 18.02g/mole  = 3.38513 moles

Total number of moles = moles of ethylene glycol + moles of water = 0.6283 + 3.38513 = 4.0134276 moles

Mole fraction = 0.6283/ 4.0134276 = 0.156549