Question

missed this? read section 6.6 (page); watch kcv 6.6 a gas mixture with a total pressure of 759 mmhg contains each of the following gases at the indicated partial pressures: 135 mmhg co₂, 229 mmhg ar, and 189 mmhg o₂. the mixture also contains helium gas. part a what is the partial pressure of the helium gas? express your answer in millimeters of mercury. part b what mass of helium gas is present in a 13.0-l sample of this mixture at 281 k? express your answer in grams.

Explanation:

Step1: Convert pressure units

First, convert the partial pressure of helium from mmHg to atm.
We know that \(1\ atm = 760\ mmHg\). So, \(P_{He}=\frac{206\ mmHg}{760\ mmHg/atm}\approx0.271\ atm\)

Step2: Use the ideal gas law \(PV = nRT\)

The ideal gas law is \(PV=nRT\), where \(P\) is pressure, \(V\) is volume (\(V = 13.0\ L\)), \(n\) is the number of moles, \(R\) is the ideal gas constant (\(R=0.0821\ L\cdot atm/(mol\cdot K)\)), and \(T\) is temperature (\(T = 281\ K\)).
We can solve for \(n\) (number of moles of \(He\)): \(n=\frac{P_{He}V}{RT}\)
Substitute the values: \(n=\frac{0.271\ atm\times13.0\ L}{0.0821\ L\cdot atm/(mol\cdot K)\times281\ K}\)

$$n=\frac{3.523}{23.0701}\ mol\approx0.1536\ mol$$

Step3: Calculate the mass of \(He\)

The molar mass of \(He\) is \(M = 4.00\ g/mol\).
The mass \(m\) of \(He\) is given by \(m=n\times M\)
Substitute \(n = 0.1536\ mol\) and \(M = 4.00\ g/mol\): \(m=0.1536\ mol\times4.00\ g/mol = 0.6144\ g\)

Answer:

\(0.614\ g\)