Question

The density of a gas is \(1.964 \mathrm{~g} \mathrm{dm}^{-3}\) at \(273 \mathrm{~K}\) and \(76 \mathrm{~cm} \mathrm{Hg}\). The gas is (a) \(\mathrm{CH}_{4}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (c) \(\mathrm{CO}_{2}\) (d) \(\mathrm{Xe}\)

Short answer

The gas is \(\mathrm{CO}_2\).

Step-by-step solution

List the Known Variables

We know the density \(d = 1.964 \mathrm{~g} \mathrm{~dm}^{-3}\), temperature \(T = 273 \mathrm{~K}\), and pressure \(P = 76 \mathrm{~cm} \mathrm{Hg}\). We also convert the pressure to atmospheres: \(76 \mathrm{~cm} \mathrm{Hg} = 1 \mathrm{~atm}\).

Apply the Ideal Gas Law to Find Molar Mass

Using the ideal gas law in the form \(PV = nRT\), where \(n = \frac{m}{M}\) (\(m\) is mass and \(M\) is molar mass), we get \(PM = dRT\). Rearrange to find \(M\):\[M = \frac{dRT}{P}\]

Insert Known Values into the Equation

Substitute the known values into the equation: \[M = \frac{1.964 \times 0.0821 \times 273}{1} = 44.02\, \mathrm{g/mol}\]

Compare Molar Mass with Options

Compare the calculated molar mass to the given options: \[\mathrm{CH}_4 = 16.04\, \mathrm{g/mol},\ \mathrm{C}_2\mathrm{H}_6 = 30.07\, \mathrm{g/mol},\ \mathrm{CO}_2 = 44.01\, \mathrm{g/mol},\ \mathrm{Xe} = 131.29\, \mathrm{g/mol}\].

Identify the Gas

The calculated molar mass \(44.02\, \mathrm{g/mol}\) closely matches \(\mathrm{CO}_2\) (\(44.01\, \mathrm{g/mol}\)). Thus, the gas is \(\mathrm{CO}_2\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of an ideal gas. It is expressed as \( PV = nRT \), where:

• \( P \) is the pressure of the gas, often measured in atmospheres (atm).
• \( V \) is the volume of the gas, typically in liters (L).
• \( n \) is the number of moles of gas.
• \( R \) is the ideal gas constant, which is \( 0.0821 \, ext{L} \cdot ext{atm/mol} \cdot ext{K} \).
• \( T \) is the temperature, measured in Kelvin (K).

This equation allows scientists and students to calculate one of these variables if the others are known. In this exercise, it's utilized with a rearrangement to find the molar mass of the gas by considering the density. By knowing the density, pressure, and temperature, you can effectively calculate properties of gases under different conditions using this powerful law.

Molar Mass Calculation

Molar mass, often referred to as molecular weight, is the mass of one mole of a chemical compound. It is expressed in grams per mole (g/mol). To calculate the molar mass using the Ideal Gas Law, we manipulate the equation into the form: \( PM = dRT \), where \( d \) is the density of the gas. Rearrange to solve for the molar mass:

• \( M = \frac{dRT}{P} \)

To effectively use this equation:- Convert all quantities to the correct units: density in \( ext{g} \cdot ext{dm}^{-3} \), temperature in Kelvin, and pressure in atmospheres.- Insert these values into the equation to find \( M \).In our exercise, inserting the known values led to the molar mass calculation of \( 44.02 \, ext{g/mol} \), which was then compared to known substances to identify the gas as carbon dioxide (\( ext{CO}_2 \)).

Density of Gases

The density of a gas is defined as its mass per unit volume. It is commonly expressed in \( ext{g} \cdot ext{dm}^{-3} \). Understanding gas density is crucial as it relates closely to the Ideal Gas Law. This law lets us connect density with other physical properties such as molar mass.In the context of the Ideal Gas Law, density plays a key role in deriving the molar mass, especially when the gas does not conform perfectly to ideal behavior.For accurate calculations:- Know that density varies with changes in temperature and pressure.- Recognize that higher density could imply a greater molar mass if other factors like pressure and temperature remain constant.In this exercise, by knowing the density of \( 1.964 \, ext{g} \cdot ext{dm}^{-3} \), it was possible to determine the gas's identity, leveraging these principles effectively in conjunction with the Ideal Gas Law.