Respuesta :
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
The mass of ethylene glycol is 184 grams.
Based on the given information,
• The vapor pressure(Vp) of pure ethanol is 102 torr.
• The vapor pressure or the final pressure (Vpf) after reducing it by 12 torr is (102 - 12) = 90 torr.
• It is known that the molar mass of ethanol is 45 g/mol, and the molar mass of ethylene glycol is 62 g/mol.
• The mass of ethanol given is 1 kg or 1000 grams.
Now by using formula,
Vp = Vpf × Xa
Here, Vp is the vapor pressure of pure solvent, Vpf is the vapor pressure of the solution, and Xa is the mole fraction of solvent.
Now putting the values we get,
[tex]90 torr = 102 torr * Xa\\Xa = \frac{90 torr }{102 torr} \\Xa = 0.88[/tex]
Now the number of moles of ethanol can be calculated by using the formula,
Moles = Given mass of ethanol/Molar mass of ethanol
[tex]Moles of ethanol = \frac{1000 g}{46 g/mol} \\Moles of ethanol = 21.74 moles[/tex]
The mole fraction can be calculated as,
[tex]Xa =\frac{na}{na+nb}[/tex]
Here na is the moles of ethanol, and nb is the moles of ethylene glycol.
Now putting the values we get,
[tex]0.88 = \frac{21.74 moles}{21.74 moles + nb} \\0.88 (21.74 moles + nb) = 21.74 moles\\19.13 moles + 0.88 nb = 21.74 moles\\0.88 nb = 2.61 moles\\nb = 2.97 moles[/tex]
Now the mass of ethylene glycol can be determined as,
nb = given mass of ethylene glycol/Molar mass of ethylene glycol
[tex]2.97 moles = \frac{Mass}{62 g/mol} \\Mass = 184 grams[/tex]
Thus, the mass of ethylene glycol is 184 grams.
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