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Chapter 5: Problem 145

A \(6.0-\mathrm{L}\) flask contains a mixture of methane \(\left(\mathrm{CH}_{4}\right),\) argon, and helium at \(45^{\circ} \mathrm{C}\) and \(1.75 \mathrm{~atm} .\) If the mole fractions of helium and argon are 0.25 and \(0.35,\) respectively, how many molecules of methane are present?

Short Answer

Expert verified
The number of methane molecules is approximately \(9.655 \times 10^{22}\).

Step by step solution

01

Convert the Temperature to Kelvin

First, the temperature needs to be converted from Celsius to Kelvin. Use the formula \(T(K) = T(°C) + 273.15\). So, \(T = 45 + 273.15 = 318.15K\)
02

Calculate the Total Number of Moles in the Flask

Use the Ideal Gas Law, \(PV = nRT\) to calculate the total number of moles \(n\). Here, \(P = 1.75 \, atm\), \(V = 6.0 \, L\), \(R = 0.0821 \, atm \, L/(mol \, K)\), and \(T = 318.15 \). Substituting these: \[n = \frac{PV}{RT} = \frac{(1.75)(6.0)}{(0.0821)(318.15)} \approx 0.401 \, mol\]
03

Calculate Moles of Helium and Argon

Using mole fractions: Number of moles of Helium \(n_{He}\) = mole fraction of He \(\times\) total moles \[ n_{He} = 0.25 \times 0.401 \approx 0.10025 \, mol\] Similarly, number of moles of Argon \(n_{Ar}\) = mole fraction of Ar \(\times\) total moles \[ n_{Ar} = 0.35 \times 0.401 \approx 0.14035 \, mol\]
04

Calculate Moles of Methane

The mole fraction of methane \(x_{CH_4}\) can be found by subtracting the mole fractions of Helium and Argon from 1. \[ x_{CH_4} = 1 - (0.25 + 0.35) = 0.40 \] Number of moles of methane \(n_{CH_4}\) = mole fraction of CH\textsubscript{4} \(\times\) total moles \[ n_{CH_4} = 0.40 \times 0.401 \approx 0.1604 \, mol \]
05

Calculate Number of Methane Molecules

Convert moles of methane to number of molecules using Avogadro's number \(6.022 \times 10^{23} \). \[\text{Number of methane molecules} = n_{CH_4} \times 6.022 \times 10^{23} \approx 0.1604 \times 6.022 \times 10^{23} \approx 9.655 \times 10^{22} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The Ideal Gas Law is a key concept in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. The equation is expressed as:
\[ PV = nRT \]
Here,
  • P represents the pressure of the gas, measured in atmospheres (atm).
  • V represents the volume of the gas, measured in liters (L).
  • n is the number of moles of gas.
  • R is the ideal gas constant, which has a value of 0.0821 atm·L/(mol·K).
  • T represents the temperature of the gas, measured in Kelvin (K).
To find the number of moles of a gas in a flask, you need to know the pressure, volume, and temperature of the gas. By rearranging the equation to solve for n, you can determine the number of moles:
\[ n = \frac{PV}{RT} \]
In our exercise, this formula helped calculate the total moles of gas in a mixture.
Mole Fraction
Mole fraction is a way to express the concentration of a component in a mixture. It is calculated by dividing the number of moles of one component by the total number of moles of all components in the mixture. For any component i in a mixture, the mole fraction (x_i) is:
\[ x_i = \frac{n_i}{n_{total}} \]
Here,
  • \(n_i\) is the number of moles of the component i.
  • \(n_{total}\) is the total number of moles in the mixture.

Mole fractions are important because they describe the proportion of each gas in the mixture. In the given exercise, we used mole fractions to find the number of moles of helium and argon. By subtracting the mole fractions of helium and argon from 1, we calculated the mole fraction of methane. This step was crucial to determine the number of moles and subsequently the number of methane molecules.
Temperature Conversion
Temperature conversions are often needed in chemistry when using equations that require the temperature in Kelvin. The temperature in Celsius (°C) can be converted to Kelvin (K) using the following formula:
\[ T(K) = T(°C) + 273.15 \]
This formula is straightforward and helps ensure accurate calculations in gas law problems. For instance, in the exercise provided, the initial temperature was given in Celsius. By converting it to Kelvin using the formula, we obtained:
\[ T = 45°C + 273.15 = 318.15 K \]
Having the temperature in Kelvin allowed us to properly apply the Ideal Gas Law, ensuring the accuracy of our calculations. Remember, always make sure your temperature is in Kelvin when using the Ideal Gas Law.

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Most popular questions from this chapter

Canadian chemists have developed a modern variation of the 1899 Mond process for preparing extremely pure metallic nickel. A sample of impure nickel reacts with carbon monoxide at \(50^{\circ} \mathrm{C}\) to form gaseous nickel carbonyl, \(\mathrm{Ni}(\mathrm{CO})_{4}\) (a) How many grams of nickel can be converted to the carbonyl with \(3.55 \mathrm{~m}^{3}\) of CO at \(100.7 \mathrm{kPa} ?\) (b) The carbonyl is then decomposed at \(21 \mathrm{~atm}\) and \(155^{\circ} \mathrm{C}\) to pure \((>99.95 \%)\) nickel. How many grams of nickel are obtained per cubic meter of the carbonyl? (c) The released carbon monoxide is cooled and collected for reuse by passing it through water at \(35^{\circ} \mathrm{C}\). If the barometric pressure is 769 torr, what volume (in \(\mathrm{m}^{3}\) ) of \(\mathrm{CO}\) is formed per cubic meter of carbonyl?

Three \(5-\mathrm{L}\) flasks, fixed with pressure gauges and small valves, each contain \(4 \mathrm{~g}\) of gas at \(273 \mathrm{~K}\). Flask A contains \(\mathrm{H}_{2}\), flask B contains He, and flask C contains \(\mathrm{CH}_{4}\). Rank the flask contents in terms of (a) pressure, (b) average kinetic energy of the particles, (c) diffusion rate after the valve is opened, (d) total kinetic energy of the particles, (e) density, and (f) collision frequency.

A person inhales air richer in \(\mathrm{O}_{2}\) and exhales air richer in \(\mathrm{CO}_{2}\) and water vapor. During each hour of sleep, a person exhales a total of about \(300 \mathrm{~L}\) of this \(\mathrm{CO}_{2}\) -enriched and \(\mathrm{H}_{2} \mathrm{O}\) -enriched air. (a) If the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) in exhaled air are each 30.0 torr at \(37.0^{\circ} \mathrm{C},\) calculate the mass \((\mathrm{g})\) of \(\mathrm{CO}_{2}\) and of \(\mathrm{H}_{2} \mathrm{O}\) exhaled in \(1 \mathrm{~h}\) of sleep. (b) How many grams of body mass does the person lose in \(8 \mathrm{~h}\) of sleep if all the \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) exhaled come from the metabolism of glucose? $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$

The gravitational force exerted by an object is given by \(F=m g\) where \(F\) is the force in newtons, \(m\) is the mass in kilograms, and \(g\) is the acceleration due to gravity \(\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) Use the definition of the pascal to calculate the mass (in \(\mathrm{kg}\) ) of the atmosphere above \(1 \mathrm{~m}^{2}\) of ocean. (b) Osmium \((Z=76)\) is a transition metal in Group \(8 \mathrm{~B}(8)\) and has the highest density of any element ( \(22.6 \mathrm{~g} / \mathrm{mL}\) ). If an osmium column is \(1 \mathrm{~m}^{2}\) in area, how high must it be for its pressure to equal atmospheric pressure? [Use the answer from part (a) in your calculation.]

Sulfur dioxide emissions from coal-burning power plants are removed by flue- gas desulfurization. The flue gas passes through a scrubber, and a slurry of wet calcium carbonate reacts with it to form carbon dioxide and calcium sulfite. The calcium sulfite then reacts with oxygen to form calcium sulfate, which is sold as gypsum. (a) If the sulfur dioxide concentration is 1000 times higher than its mole fraction in clean, dry air \(\left(2 \times 10^{-10}\right),\) how much calcium sulfate \((\mathrm{kg})\) can be made from scrubbing \(4 \mathrm{GL}\) of flue gas ( \(\left.1 \mathrm{GL}=1 \times 10^{9} \mathrm{~L}\right)\) ? A state-of-the-art scrubber removes at least \(95 \%\) of the sulfur dioxide. (b) If the mole fraction of oxygen in air is \(0.209,\) what volume \((\mathrm{L})\) of air at \(1.00 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) is needed to react with all the calcium sulfite?
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