Answer:
183.7 g of ethylene glycol
Explanation:
This a case of colligative property, about vapour pressure lowering.
ΔPv = P° . Xst
ΔPv = 12 Torr
P° = Vapour pressure of pure solvent → 100 Torr
12 Torr = 100 Torr . Xst
12 Torr / 100 Torr = 0.12 → Xst
Xst is mole fraction for solute which means moles of solute / Total moles and, the total moles are the sum of moles of solvent + moles of solute.
We can make this equation:
Moles of solute / (Moles of solute + Moles of solvent) = 0.12
We don't know the moles of solute, but we do know the moles of solvent by the mass.
Mass / Molar mass = Mol
1000 g / 46 g/m = 21.7 moles
Moles of solute / Moles of solute + 21.7 moles = 0.12
Moles of solute = 0.12 ( Moles of solute + 21.7 moles)
Moles of solute - 0.12 moles of solute = 2.60 moles
0.88 moles of solute = 2.60 moles
Moles of solute = 2.60 moles / 0.88 → 2.96 moles of solute
Mol . molar mass = Mass
2.96 m . 62g/m = 183.7 g
