Problem 8 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

At \(300 \mathrm{~K}\), the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen \(\left(\mathrm{N}_{2}\right)\) at 4 bar. The molar mass of gaseous molecule is : (a) \(28 \mathrm{~g} \mathrm{~mol}^{-1}\) (b) \(56 \mathrm{~g} \mathrm{~mol}^{-1}\) (c) \(112 \mathrm{~g} \mathrm{~mol}^{-1}\) (d) \(224 \mathrm{~g} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The molar mass of the gaseous molecule is 112 g/mol (option c).

Step 1: Apply the Ideal Gas Law

The Ideal Gas Law is expressed as \(PV = nRT\). Using it, the density \(\rho\) of a gas can be linked to its molar mass \(M\) via \(\rho = \frac{PM}{RT}\). Here, \(P\) is the pressure, \(R\) is the gas constant, and \(T\) is the temperature.

Step 2: Set up the Density Ratio

Let the molar mass of the gaseous molecule be \(M'\) and for nitrogen it is \(M = 28 \, \text{g/mol}\). Given that \(\rho' = 2\rho_{\text{N}_2}\) and using the density formula from Step 1:\[\frac{P'M'}{RT} = 2 \times \frac{PM}{RT}\]Simplifying gives:\[M' = 2 \times \frac{PM}{P'}\]

Step 3: Plug in Known Values

Substitute \(P = 4 \, \text{bar}\), \(P' = 2 \, \text{bar}\), and \(M = 28 \, \text{g/mol}\) into the equation from Step 2:\[M' = 2 \times \frac{4 \, \text{bar} \times 28 \, \text{g/mol}}{2 \, \text{bar}}\]

Step 4: Simplify the Expression

Simplify the expression to find the molar mass:\[M' = 2 \times 56 \, \text{g/mol} = 112 \, \text{g/mol}\]

Step 5: Verify with Given Options

Compare the result with the choices:(a) 28 \, \text{g mol}^{-1}(b) 56 \, \text{g mol}^{-1}(c) 112 \, \text{g mol}^{-1}(d) 224 \, \text{g mol}^{-1}The calculated molar mass \(112 \, \text{g/mol}\) matches option (c).

Key Concepts

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry. It's essentially the mass of one mole of a substance, usually expressed in grams per mole (g/mol). Understanding how to calculate it helps determine the amount of a substance in a sample. When dealing with gases, the Ideal Gas Law comes in handy: \[ PV = nRT \] Here's a simple breakdown:- **P** is the pressure (in atmospheres or bars).- **V** is the volume, though you often end up focusing on density instead.- **n** is the number of moles.- **R** is the gas constant.- **T** is the temperature (in Kelvin). We can derive a useful formula for density: \(\rho = \frac{PM}{RT}\). Here, density (\(\rho\)) relates directly to pressure (\(P\)) and molar mass (\(M\)). By manipulating this formula, you can easily find the molar mass given pressure and density values. In the example problem, setting up a density ratio helped uncover the molar mass of the unknown gas.

Density of Gases

Density plays a crucial role in understanding gases and their behavior under different conditions. Defined as mass per unit volume, density is especially important in contexts like this exercise, where gases are involved. Using the density\( \rho = \frac{M \cdot P}{R \cdot T} \) formula derived from the Ideal Gas Law, you see how pressure, temperature, and molar mass influence the density of a gas. In the context of comparing two gases, like the unknown gas and nitrogen in this exercise, setting a density ratio becomes quite revealing. For instance, when it’s said that the density of the unknown gas is twice that of nitrogen under given conditions, this information can be used with the density formula to discover the unknown gas's molar mass. It's all about setting reasonable comparisons or equalities based on provided data and the behavior described by gas laws.

Pressure and Temperature Relationship

Pressure and temperature are tightly interwoven in the study of gas laws. One fundamental concept here is that at a constant volume, the pressure of a gas is directly proportional to its temperature (measured in Kelvin), as described by Gay-Lussac's Law. This relationship is part of the combined gas law, expressed as \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), under consistent volume. However, in our problem scenario, it's essential to observe the impact of pressure changes on density and subsequently on calculations like molar mass. When these relationships are considered, you can effectively use them to unravel unknown variables. Manipulating scenarios by shifting pressures (while keeping temperature constant) can reveal about molecules like their molar mass, as shown when doubling the density by halving the initial pressure, all at the same temperature.