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(a) Calculate the density of \(\mathrm{NO}_{2}\) gas at \(0.970 \mathrm{~atm}\) and \(35^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a gas if \(2.50 \mathrm{~g}\) occupies \(0.875 \mathrm{~L}\) at 685 torr and \(35^{\circ} \mathrm{C}\).

Short Answer

Expert verified
a) The density of NO鈧 gas at 0.970 atm and 35掳C is approximately \(1.77 \mathrm{g/L}\). b) The molar mass of the gas, given that 2.50 g occupies 0.875 L at 685 torr and 35掳C, is approximately \(68.87 \mathrm{g/mol}\).

Step by step solution

01

Convert temperature to Kelvin

We need to make sure the temperature is measured in Kelvin for the gas law calculations. To convert Celsius to Kelvin, add 273.15 to the given temperature: \(T_K = T_C + 273.15\) \(T_K = 35 + 273.15 = 308.15 \mathrm{K}\)
02

Calculate the molar mass of NO鈧

To calculate the molar mass of NO鈧, we add the molar masses of one nitrogen atom and two oxygen atoms: Molar mass of N = 14.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of NO鈧 = 14.01 + 2(16.00) = 46.01 g/mol
03

Use the Ideal Gas Law to calculate the density

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 ideal gas constant (0.0821 L atm/mol K), and T is the temperature. We'll modify the equation to solve for density (蟻), which is given by \(\rho =\frac{mass}{volume}\): \(\rho = \frac{n \times Molar\: mass}{V}\) Replace n/V from Ideal Gas Law equation, \(\frac{n}{V} = \frac{P}{RT}\) So, \(\rho = \frac{P \times Molar\:mass}{R \times T}\) Now plug in the values: \(\rho = \frac{0.970 \mathrm{atm} \times 46.01 \mathrm{g/mol}}{0.0821 \mathrm{L \times atm/mol \times K} \times 308.15 \mathrm{K}}\) After calculating, we get the density: \(\rho \approx 1.77 \mathrm{g/L}\) b) Calculate the molar mass of a gas if 2.50 g occupies 0.875 L at 685 torr and 35掳C.
04

Convert pressure to atm and temperature to Kelvin

We need to make sure the pressure is measured in atm and temperature is measured in Kelvin for the gas law calculations: Pressure in atm = \(685 \mathrm{torr} \times \frac{1 \mathrm{atm}}{760 \mathrm{torr}} \approx 0.901 \mathrm{atm}\) Temperature in Kelvin, \(T_K = T_C + 273.15\) \(T_K = 35 + 273.15 = 308.15 \mathrm{K}\)
05

Calculate number of moles of the gas

Again, use the Ideal Gas Law equation, \(PV=nRT\), we'll solve for n: n = \(\frac{PV}{RT}\) Now plug in the values: \(n = \frac{0.901 \mathrm{atm} \times 0.875 \mathrm{L}}{0.0821 \mathrm{L \times atm/mol \times K} \times 308.15 \mathrm{K}}\) After calculating, we get the number of moles: n 鈮 0.0363 mol
06

Calculate the molar mass of the gas

Since we have the mass and moles of the gas, we can calculate its molar mass: Molar mass = \(\frac{mass}{moles}\) Molar mass = \(\frac{2.50 \mathrm{g}}{0.0363 \mathrm{mol}}\) After calculating, we get the molar mass: Molar mass 鈮 68.87 g/mol

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Key Concepts

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

Gas Density Calculation
Calculating the density of a gas involves using the Ideal Gas Law, which is crucial in understanding how various properties of gases are interrelated.
To find the density (\( \rho \)) of a gas, we utilize a rearranged form of the Ideal Gas Law.
The key relationship here is \( \rho = \frac{P \times \text{Molar mass}}{R \times T} \). In this formula:
  • \( P \) represents the pressure of the gas.
  • \( R \) is the gas constant, which is approximately \( 0.0821 \, \text{L atm/mol K} \).
  • \( T \) stands for the temperature in Kelvin, which is important for accuracy.
  • The pressure and temperature of the gas, along with its molar mass, allow us to determine its density. By plugging in these values:
    • Pressure = 0.970 atm
    • Molar Mass = 46.01 g/mol
    • Temperature = 308.15 K
    we find that the density of \( \text{NO}_2 \) gas is approximately 1.77 g/L.
    This calculation is a fundamental type for scenarios where pressure, molar mass, and temperature are given, allowing one to determine the density directly from these properties.
Molar Mass Determination
Understanding how to determine the molar mass of gases is a significant concept in chemistry. To calculate the molar mass, you first need to know the mass and the number of moles of the gas.
We derive the number of moles (\( n \)) using the Ideal Gas Law, \( PV = nRT \), rearranged to solve for \( n \).
For example, with a 2.50 g gas occupying 0.875 L at 685 torr and 35 掳C, first convert the pressure to atm (\( \approx 0.901 \) atm) and the temperature to Kelvin (308.15 K):
  • \( n = \frac{PV}{RT} = \frac{0.901 \, \text{atm} \times 0.875 \, \text{L}}{0.0821 \, \text{L atm/mol K} \times 308.15 \, \text{K}} \approx 0.0363 \, \text{mol} \)
Then, the molar mass is calculated as \( \text{Molar mass} = \frac{\text{mass}}{\text{moles}} \), which results in approximately 68.87 g/mol for this gas sample.
This approach provides a clear method to discover the molar mass of gases when their physical properties are known.
Conversion of Units in Gas Laws
Proper conversion of units is necessary for any gas law calculations to ensure accuracy and consistency, particularly when determining gas properties like density or molar mass.
Common conversions include:
  • Temperature from Celsius to Kelvin: \( T_K = T_C + 273.15 \)
  • Pressure from torr to atm: Utilize the conversion factor \( \frac{1 \, \text{atm}}{760 \, \text{torr}} \)
Temperature must always be in Kelvin because gas laws use absolute temperature to maintain correct and proportional relationships.Similarly, pressure should be converted to atm if using the gas constant \( R = 0.0821 \, \text{L atm/mol K} \).
Through proper unit conversion, calculations based on the Ideal Gas Law, such as gas density and molar mass, will be precise and can be reliably used to derive meaningful chemical insights.

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Most popular questions from this chapter

A neon sign is made of glass tubing whose inside diameter is \(2.5 \mathrm{~cm}\) and whose length is \(5.5 \mathrm{~m}\). If the sign contains neon at a pressure of 1.78 torr at \(35^{\circ} \mathrm{C}\), how many grams of neon are in the sign? (The volume of a cylinder is \(\pi r^{2} h\).)

A fixed quantity of gas at \(21^{\circ} \mathrm{C}\) exhibits a pressure of 752 torr and occupies a volume of 5.12 L. (a) Calculate the volume the gas will occupy if the pressure is increased to 1.88 atm while the temperature is held constant. (b) Calculate the volume the gas will occupy if the temperature is increased to \(175^{\circ} \mathrm{C}\) while the pressure is held constant.

(a) How is the law of combining volumes explained by Avogadro's hypothesis? (b) Consider a 1.0 - \(\mathrm{L}\) flask containing neon gas and a 1.5-L flask containing xenon gas. Both gases are at the same pressure and temperature. According to Avogadro's law, what can be said about the ratio of the number of atoms in the two flasks? (c) Will 1 mol of an ideal gas always occupy the same volume at a given temperature and pressure? Explain.

Rank the following gases from least dense to most dense at 1.00 atm and \(298 \mathrm{~K}: \mathrm{SO}_{2}, \mathrm{HBr}, \mathrm{CO}_{2} .\) Explain.

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathrm{a})\) If \(1.56 \mathrm{~g}\) of cyclopropane has a volume of \(1.00 \mathrm{~L}\) at 0.984 atm and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

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