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Chapter 10: Problem 117

Dry nitrogen gas is bubbled through liquid benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right.\) ) at \(20.0^{\circ} \mathrm{C}\). From \(100.0 \mathrm{~L}\) of the gaseous mixture of nitrogen and benzene, \(24.7 \mathrm{~g}\) benzene is condensed by passing the mixture through a trap at a temperature where nitrogen is gaseous and the vapor pressure of benzene is negligible. What is the vapor pressure of benzene at \(20.0^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The vapor pressure of benzene at \(20.0^{\circ}C\) is 0.0762 atm.

Step by step solution

01

Use Ideal Gas Law to find the moles of benzene

We know that the Ideal Gas Law is given by the formula: \(PV=nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin using the following formula: \(T(K) = T(°C) + 273.15\) \(T(K) = 20.0 + 273.15 = 293.15 \,K\) Next, we can use the mass of benzene and its molar mass to find the moles of benzene: moles of benzene = \(\frac{\text{mass}}{\text{molar mass}}\) moles of benzene = \(\frac{24.7 \,g}{78.11 \,g/mol} = 0.316 \,mol\)
02

Calculate the partial pressure of benzene

Now that we know the moles of benzene in the gaseous mixture, we can rearrange the Ideal Gas Law formula to solve for the pressure: \(P_{benzene} = \frac{n_{benzene}RT}{V}\) \(P_{benzene} = \frac{0.316 \,mol \times 0.0821 \,L \cdot atm \, mol^{-1} \, K^{-1} \times 293.15 \,K}{100 \,L}\) \(P_{benzene} = 0.0762 \,atm\)
03

Calculate the vapor pressure of benzene

Since the partial pressure of benzene in the gaseous mixture is equal to its vapor pressure at the given temperature, the vapor pressure of benzene at 20°C is: Vapor pressure of benzene = \(P_{benzene} = 0.0762 \, atm\) The vapor pressure of benzene at 20°C is 0.0762 atm.

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