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Chapter 6: Problem 101

What is the partial pressure of \(\mathrm{Cl}_{2}(\mathrm{g}),\) in millimeters of mercury, at \(0.00^{\circ} \mathrm{C}\) and 1.00 atm in a gaseous mixture that consists of \(46.5 \% \mathrm{N}_{2}, 12.7 \% \mathrm{Ne},\) and \(40.8 \%\) \(\mathrm{Cl}_{2},\) by mass?

Short Answer

Expert verified
The partial pressure of \(Cl_{2}(g)\) is 423.5mmHg

Step by step solution

01

Calculate the mass of \(Cl_2\)

Assume the total mass of the gaseous mixture to be 100g. Using the given percentage of \(Cl_2\) which is 40.8%, calculate the mass of \(Cl_2\) in the gaseous mixture as: \(40.8g\).
02

Determine the number of moles of \(Cl_2\)

With the molecular mass of \(Cl_2\) being approximately 71 g/mol, the number of moles of \(Cl_2\) can be calculated as: \(40.8g / 71g/mol = 0.574 mol\)
03

Use the Ideal Gas Law

The Ideal Gas Law is given by \(PV = nRT\), 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. Solving the equation for \(P\), we get \(P = nRT/V\). The values for \(V\), \(R\), and \(T\) are not given but we know from Standard Ambient Temperature and Pressure (SATP) conditions that \(V=22.4L\), \(R=62.4L.mmHg/mol.K\), and \(T=273.15 K\). Substituting these values into the equation, we calculate the total pressure of the gas mix as \(P=nRT/V = 0.574mol*(62.4L.mmHg/mol.K)*(273.15K)/22.4L = 738.5 mmHg\)
04

Calculate the partial pressure of \(Cl_2\)

The partial pressure of a component in a gaseous mixture is the portion of the total pressure contributed by that component. It is equal to the total pressure times the mole fraction of the component. The molar fraction of \(Cl_2\) is \(0.574mol/1.00mol = 0.574\). Substituting these values into the equation for partial pressure, we get the partial pressure of \(Cl_2 = 738.5mmHg*0.574 = 423.5mmHg\)

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