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A sample of methane (CH4) gas contains a small amount of

Chapter 5, Problem 5.62

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QUESTION:

A sample of methane \(\left(\mathrm{CH}_{4}\right)\) gas contains a small amount of helium. Calculate the volume percentage of helium if the density of the sample is 0.70902 g/L at \(0.0^{\circ} \mathrm{C}\) and 1.000 atm.

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QUESTION:

A sample of methane \(\left(\mathrm{CH}_{4}\right)\) gas contains a small amount of helium. Calculate the volume percentage of helium if the density of the sample is 0.70902 g/L at \(0.0^{\circ} \mathrm{C}\) and 1.000 atm.

ANSWER:

Step 1 of 3

Methane is the simplest organic compound, the simplest form of alkane, and natural gas main constituent.

By using the density of the mixed gas sample, we can calculate the average molecular mass for the sample by using the formula given below:

\(\text { Molar mass }=\frac{d \times R \times T}{P}\);

Where,

*d is the density of the sample, i.e., 0.70902 g/L.

*R is the universal gas constant, i.e., \(0.08206\mathrm{\ L}\cdot\mathrm{atm}\cdot\mathrm{K}^{-1}\cdot\mathrm{mol}^{-1}\).

*P is the pressure of the gas, i.e., 1.00 atm.

*T is the temperature of the gas, i.e., \(0.08^{\circ}\mathrm{C}(0.08+273.15)K=273.23\mathrm{\ K}\).

On substituting all the known values in the above equation, we get

\(\begin{aligned}\text{ Molar mass }&=\frac{d\times R\times T}{P}\\ &=\frac{0.70902\mathrm{\ g}/\mathrm{L}\times0.08206\mathrm{\ L}\cdot\mathrm{atm}\cdot\mathrm{K}^{-1}\cdot\mathrm{mol}^{-1}\times273.23\mathrm{\ K}}{1.00\mathrm{\ atm}}\\ &=15.897\mathrm{\ g}/\mathrm{mol}\end{aligned}\) 

This value represents the weighted average of the molecular masses of the two components, methane and helium.

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