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A sample of gas has a mass of \(38.8 \mathrm{mg}\). Its volume is \(224 \mathrm{~mL}\) at a temperature of \(55^{\circ} \mathrm{C}\) and a pressure of 886 torr. Find the molar mass of the gas.

Short Answer

Expert verified
The molar mass of the gas is 4.69 g/mol.

Step by step solution

01

Convert the temperature to Kelvin

The temperature must be in Kelvin for gas law calculations. Convert Celsius to Kelvin using the formula: Kelvin = Celsius + 273.15. Therefore, the temperature in Kelvin is: 55°C + 273.15 = 328.15 K.
02

Convert the pressure to atmospheres

Since the Ideal Gas Law uses pressure in atmospheres, convert the pressure from torr to atmospheres. 1 atm = 760 torr. So, the pressure in atmospheres is: 886 torr * (1 atm / 760 torr) = 1.166 atm.
03

Convert the mass to grams

The mass should be in grams to find the molar mass. Convert milligrams to grams: 38.8 mg * (1 g / 1000 mg) = 0.0388 g.
04

Use the Ideal Gas Law

Apply the Ideal Gas Law, PV = nRT, where P is pressure, V is volume, R is the gas constant (0.0821 L atm/mol K), and T is temperature in Kelvin. We solve for n, the number of moles, as we have P, V and T.
05

Calculate the number of moles (n)

Rearrange the Ideal Gas Law to solve for n: n = PV / RT. Plugging in the values: n = (1.166 atm * 224 mL * (1 L / 1000 mL)) / (0.0821 L atm/mol K * 328.15 K) = 0.00826 moles.
06

Find the molar mass

The molar mass (M) is the mass of the gas divided by the number of moles: M = mass / n. Therefore, M = 0.0388 g / 0.00826 moles = 4.69 g/mol.

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Key Concepts

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

Ideal Gas Law
When dealing with gases, a fundamental concept to understand is the Ideal Gas Law, represented by the equation: \[ PV = nRT \]. Here, \(P\) stands for the pressure of the gas, \(V\) indicates its volume, \(n\) is the number of moles of gas present, \(R\) is a constant known as the universal gas constant, and \(T\) is the temperature of the gas measured in Kelvin. This law is crucial because it relates the four state variables (P, V, n, and T) for a sample of an ideal gas.

In practice, this means that if you are given any three of these state variables, you can use the Ideal Gas Law to calculate the fourth. For instance, if we know the pressure, the volume, and the temperature of a gas, we can determine the amount of gas in moles. It's important to recognize that real gases often deviate from ideal behavior under high pressure or low temperature, but for many conditions, particularly those involving common temperature and pressure ranges, the Ideal Gas Law can approximate their behavior effectively.
Converting Temperature to Kelvin
Temperature measurements for gas law calculations must be in Kelvin, as it is the SI unit for thermodynamic temperature and is used in various scientific calculations. To convert Celsius to Kelvin, you utilize the formula: \[ \text{Kelvin} = \text{Celsius} + 273.15 \].

This straightforward addition accounts for the fact that 0 Kelvin, or absolute zero (the theoretical point where particles have minimal vibrational motion), is equivalent to -273.15°C. Knowing how to correctly convert temperatures into Kelvin is vital for not only using the Ideal Gas Law but for many other scientific computations as well.
Converting Pressure to Atmospheres
Pressure is another critical variable in gas law calculations, and it can be measured in various units, such as atmospheres (atm), torr, or pascals (Pa). In the context of the Ideal Gas Law, pressure is typically referenced in atmospheres. To convert from torr to atmospheres, the conversion factor used is based on the equivalence of 1 atm to 760 torr: \[ 1 \text{ atm} = 760 \text{ torr} \].

To convert torr to atm, you would divide the number of torr by 760. Accurate conversion ensures precise calculations, which is essential for any scientific work or homework problem involving gas laws.
Moles Calculation
The concept of moles is central to chemistry, as it is a unit of measurement for substance quantity. One mole corresponds to Avogadro's number of particles (atoms, molecules, ions, etc.), which is approximately \(6.022 \times 10^{23}\). Calculating the number of moles \(n\) of a substance is a common requirement in chemistry problems.

From the Ideal Gas Law, the moles of a gas can be found using the formula: \[ n = \frac{PV}{RT} \], where pressure (P) and volume (V) must be in the correct units, atmospheres and liters respectively, and temperature (T) in Kelvin. The gas constant (R) has a value of \(0.0821 \text{L atm/mol K}\). Ultimately, to find the molar mass (M) of the substance, you divide the mass of the sample by the number of moles calculated: \[ M = \frac{\text{mass}}{n} \]. This final step is often used to identify or confirm the identity of a substance based on its molar mass.

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Most popular questions from this chapter

In a common classroom demonstration, a balloon is filled with air and drenched with liquid nitrogen. The balloon contracts as the gases within the balloon cool. Suppose a balloon initially contains \(2.95 \mathrm{~L}\) of air at a temperature of \(25.0^{\circ} \mathrm{C}\) and a pressure of 0.998 atm. Calculate the expected volume of the balloon upon cooling to \(-196^{\circ} \mathrm{C}\) (the boiling point of liquid nitrogen). When the demonstration is carried out, the actual volume of the balloon decreases to 0.61 L. How does the observed volume of the balloon compare to your calculated value? Explain the difference.

Two identical balloons are filled to the same volume, one with air and one with helium. The next day, the volume of the airfilled balloon has decreased by \(5.0 \% .\) By what percent has the volume of the helium-filled balloon decreased? (Assume that the air is four-fifths nitrogen and one-fifth oxygen and that the temperature did not change.)

An automobile tire has a maximum rating of 38.0 psi (gauge pressure). The tire is inflated (while cold) to a volume of \(11.8 \mathrm{~L}\) and a gauge pressure of 36.0 psi at a temperature of \(12.0^{\circ} \mathrm{C} .\) On a hot day, the tire warms to \(65.0^{\circ} \mathrm{C}\), and its volume expands to \(12.2 \mathrm{~L} .\) Does the pressure in the tire exceed its maximum rating? (Note: The gauge pressure is the difference between the total pressure and atmospheric pressure. In this case, assume that atmospheric pressure is 14.7 psi.)

Calculate the ratio of effusion rates for Ar and \(\mathrm{Kr}\).

Consider a 1.0-L sample of helium gas and a 1.0-L sample of argon gas, both at room temperature and atmospheric pressure. a. Do the atoms in the helium sample have the same average kinetic energy as the atoms in the argon sample? b. Do the atoms in the helium sample have the same average velocity as the atoms in the argon sample? c. Do the argon atoms, because they are more massive, exert a greater pressure on the walls of the container? Explain. d. Which gas sample has the faster rate of effusion?

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