An aqueous antifreeze solution is 40.0% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.75 g/cm3. Calculate the molality, molarity, and mole fraction of the ethylene glycol.

Xm ethylene glycol = 0.16

Molality = 10.75 m

Molarity = 11.3M

Explanation:

40.0% by mass. This data says that 40 g of ethyelen glycol are contained in 100 g of solution

Therefore: Mass of solute is 40g

Mass of solvent = 60 g

Mass of solution = 100 g

We convert the mass to moles → 40 g . 1mol / 62 g = 0.645 moles

We convert the solvent mass to moles → 60 g . 1mol /18g = 3.33 moles

Xm ethylene glycol = Moles of ethylene glycol / Total moles

0.645 / (0.645 + 3.33) = 0.16

Molality are the moles of solute in 1kg of solvent.

We convert the mass of solvent to kg → 60 g . 1kg/1000g = 0.060kg

Molality (mol/kg) = 0.645 mol /0.060kg = 10.75 m

To determine molarity we use the density of solution in order to find out the volume: density = mass / volume

1.75 g/cm³ = 100 g / volume → 100 g / 1.75 g/cm³ = 57.1 cm³  (1cm³ = 1ml)

We convert the solution's volume to L → 57.1 mL . 1L/1000mL = 0.0571L

Molarity → moles of solute in 1L of solution

Molarity (mol/L) → 0.645 mol / 0.0571L = 11.3M

2b2t

explanation:

2b2t

the answer would be a