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At 1 atm and \(273 \mathrm{~K}\) the density of gas, whose molecular weight is 45 , is: (a) \(44.8 \mathrm{~g} / \mathrm{L}\) (b) \(11.4 \mathrm{~g} / \mathrm{L}\) (c) \(2 \mathrm{~g} / \mathrm{L}\) (d) \(3 \mathrm{~g} / \mathrm{L}\)

Short Answer

Expert verified
The density of the gas at 1 atm and 273 K with a molecular weight of 45 is approximately 2 g/L.

Step by step solution

01

Note the Given Parameters

The problem provides the temperature as 273 K, pressure as 1 atm, and the molecular weight of the gas as 45 g/mol.
02

Use the Ideal Gas Law

Use the Ideal Gas Law equation, \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature. Since we need the density, rearrange the equation to find \(n/V\) (moles per volume), which is equal to the density times the molecular weight. The equation becomes \(P = (\text{density} \times \text{molecular weight}) \times R \times T\).
03

Calculate the Density

Now plug in the values to calculate the density. The ideal gas constant \(R\) in units of L atm/K mol is approximately 0.0821. The pressure \(P\) is 1 atm, the molecular weight is 45 g/mol, and the temperature \(T\) is 273 K. The density is therefore calculated as follows: \( \text{density} = \frac{P}{\text{molecular weight} \times R \times T}\). Calculate the numerator 1 atm and the denominator \(45 \times 0.0821 \times 273\) to find the density in g/L.
04

Simplify the Calculation and Determine the Density

Perform the calculation to find the density as \( \text{density} = \frac{1}{45 \times 0.0821 \times 273}\). After simplifying, we find that the density is approximately 2.0 g/L.

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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 the cornerstone of gas calculations, including those involving gas density. It's an equation of state that relates pressure (\( P \)), volume (\( V \)), temperature (\( T \)), and number of moles (\( n \)) for an ideal gas. Represented as \( PV = nRT \), where \( R \) is the gas constant, this law assumes that gas particles do not interact and that the volume occupied by gas molecules is negligible.

When working with the Ideal Gas Law, remember it's crucial for all units to be consistent. In the context of density calculations, if pressure is in atmospheres (atm) and volume is in liters (L), the value of \( R \) should be \( 0.0821 \text{L atm K}^{-1} \text{mol}^{-1} \). Knowing how to manipulate this formula is essential for solving a wide array of problems in chemistry, including finding the density of a gas under given conditions.
Molecular Weight
Molecular weight, also known as molecular mass, is the sum of the atomic masses of all the atoms in a molecule. It's measured in atomic mass units (amu) or grams per mole (g/mol). In the context of gas calculations, molecular weight allows us to relate moles to mass, which is particularly handy when using the Ideal Gas Law.

Molecular weight is critical to computing the number of moles from a given mass or vice versa. For instance, if we know the molecular weight of a substance and we have a certain mass of it, we can easily calculate the number of moles by dividing the mass by the molecular weight. This step paves the way for a plethora of calculations involving the moles of a substance, including those needed for density calculations.
Gas Density
Understanding gas density is essential when working with gases in various chemical applications. Gas density is the mass of the gas per unit volume and is usually expressed in grams per liter (g/L). Unlike liquids and solids, the density of a gas is greatly affected by pressure and temperature, hence it's often calculated using the Ideal Gas Law.

To find the density of a gas, the Ideal Gas Law is rearranged to express moles per volume (\( n/V \)), which equates to density (\( \rho \times \text{molecular weight} \times R \times T \times P \times 1 \tau\1 \times 82.1 \times 73 \times 45 \times 02.1 \times 73\text{g/L}\)). Thus, gas density allows chemists to predict how a gas will behave under changing pressure and temperature conditions, especially in stoichiometry and reaction calculations.
JEE Physical Chemistry Problems
The Joint Entrance Examination (JEE) is a highly competitive test for aspiring engineers in India. Physical Chemistry forms a significant portion of the JEE Chemistry syllabus, where a thorough understanding of concepts like the Ideal Gas Law, molecular weights, and gas densities is tested. Problems related to these concepts are common and often integrate theoretical knowledge with practical problem-solving skills.

Effective preparation for JEE Physical Chemistry problems includes practicing a variety of problems, understanding the underlying principles, and being able to apply concepts in new and unfamiliar situations. Familiarity with formulas and their manipulations, like those used in density calculations, is incredibly important. Students should concentrate on understanding each step involved in the solution to deepen their conceptual knowledge and enhance their ability to tackle any question in the exam efficiently.

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