Question

(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

a) The density of NO₂ gas at 0.970 atm and 35°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°C, is approximately \(68.87 \mathrm{g/mol}\).

Step-by-step solution

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: \(T_K = T_C + 273.15\) \(T_K = 35 + 273.15 = 308.15 \mathrm{K}\)

Calculate the molar mass of NO₂

To calculate the molar mass of NO₂, 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 NO₂ = 14.01 + 2(16.00) = 46.01 g/mol

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 (ρ), 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°C.

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, \(T_K = T_C + 273.15\) \(T_K = 35 + 273.15 = 308.15 \mathrm{K}\)

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 ≈ 0.0363 mol

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 ≈ 68.87 g/mol

Key concepts

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 °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.