Answer :
78.3 Torr would be the vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 k
This question can be answered using Raoult's law and Dalton's law of partial pressure.
Raoult's law: Raoult postulated a mathematical relation between partial pressure and mole fraction. This connection is known as Raoult's Law, which asserts that the partial vapour pressure of each volatile component in a solution (obtained by dissolving a solute in a solvent) is directly proportionate to its mole fraction. The formula can be given by:
p₁= p₁⁰x₁ p₂=p₂⁰x₂
Here x denotes the mole fraction
p⁰ denotes the pressure in pure state.
Dalton's law: Dalton's law of partial pressure states that a gas mixture's total pressure is equal to the sum of its component gases' partial pressures. The pressure that each gas would produce if it filled the same volume of the mixture by itself at the same temperature is known as the partial pressure.
Mathematically,
p= p₁ + p₂
here p is the total pressure.
Now according to the given question
given mass of ethanol= 80 gm
moar mass of ethanol= 46 gm/mol
no. of moles= [tex]\frac{given mass}{molar mass} = \frac{80}{46} = 1.7 moles[/tex]
given mass of methanol= 97gm
molar mass of methanol= 32 gm/mol
no. of moles= [tex]\frac{given mass}{molar mass} = \frac{97}{32} = 3 moles[/tex] (approx.)
total number of moles= 1.7+3= 4.7 moles
mole fraction x for ethanol= [tex]\frac{1.7}{4.7} =0.36[/tex]
mole fraction of methanol= 1-0.36= 0.64
given pressure of ethanol in pure state= 97.7 Torr
given pressure of methanol in pure state= 44.6 Torr
Now from dalton's ;aw
total pressure is the sum of product of partial fraction and pressure in pure state for individual liquids.
P(total)= 0.36×44.6+ 0.64× 97.7= 78.3 Torr
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Final answer:
To find the vapor pressure of a mixture of methanol and ethanol at 293 K, we can use Raoult's law. The vapor pressure of the mixture is equal to the mole fraction of methanol multiplied by the vapor pressure of methanol plus the mole fraction of ethanol multiplied by the vapor pressure of ethanol. In this case, the vapor pressure of the mixture is 75.3 torr.
Explanation:
To find the vapor pressure of a mixture of methanol and ethanol at 293 K, we can use Raoult's law. Raoult's law states that the vapor pressure of a component in a mixture is directly proportional to its mole fraction in the mixture.
First, we need to calculate the mole fraction of each component. The mole fraction of ethanol can be calculated by dividing the moles of ethanol by the total moles of both compounds.
Next, we can use the mole fraction of each component to calculate the vapor pressure of the mixture. The vapor pressure of the mixture is equal to the mole fraction of methanol multiplied by the vapor pressure of methanol plus the mole fraction of ethanol multiplied by the vapor pressure of ethanol.
In this case, the mole fraction of methanol is 97 g / (32 g/mol) / (97 g / (32 g/mol) + 80 g / (46 g/mol)) = 0.639 and the mole fraction of ethanol is 80 g / (46 g/mol) / (97 g / (32 g/mol) + 80 g / (46 g/mol)) = 0.361. Substituting these values into the equation, we get the vapor pressure of the mixture = (0.639 * 97.7 torr) + (0.361 * 44.6 torr) = 75.3 torr. Therefore, the vapor pressure of the mixture is 75.3 torr.
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