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The vapor pressure of pure water at 55°C is 118 mm Hg. The vapor pressure of ethanol at 55°C is 144 mm Hg. What is the total vapor pressure above a mixture of 77.0 g of ethanol (CH3CH2OH, molar mass = 46.0 g/mol) and 32.0 g of water?

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User Jeruki
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1 Answer

6 votes

Answer:

To calculate the total vapor pressure above a mixture of two liquids, we can use Raoult's Law, which states that the partial vapor pressure of each component in a mixture is proportional to its mole fraction in the mixture.

The formula for calculating the partial vapor pressure of a component in a mixture is:

Partial Vapor Pressure = Mole Fraction * Vapor Pressure of Pure Component

Given:

- Vapor pressure of pure water (P₁) = 118 mm Hg

- Vapor pressure of ethanol (P₂) = 144 mm Hg

- Mass of ethanol (m₁) = 77.0 g

- Molar mass of ethanol (M₁) = 46.0 g/mol

- Mass of water (m₂) = 32.0 g

- Molar mass of water (M₂) = 18.0 g/mol

First, calculate the moles of each component:

Moles of ethanol (n₁) = m₁ / M₁

Moles of water (n₂) = m₂ / M₂

Calculate the total moles of the mixture:

Total moles (n_total) = n₁ + n₂

Calculate the mole fractions:

Mole fraction of ethanol (χ₁) = n₁ / n_total

Mole fraction of water (χ₂) = n₂ / n_total

Now, use Raoult's Law to calculate the partial vapor pressures:

Partial Vapor Pressure of ethanol (P₁') = χ₁ * P₁

Partial Vapor Pressure of water (P₂') = χ₂ * P₂

Finally, the total vapor pressure above the mixture is the sum of the partial vapor pressures:

Total Vapor Pressure = P₁' + P₂'

Substitute the given values and calculated values to find the total vapor pressure above the mixture.

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User Higty
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