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\(20 \mathrm{~g}\) of an ideal gas contains only atoms of \(\mathrm{S}\) and \(\mathrm{O}\) occupies \(5.6 \mathrm{~L}\) at \(1 \mathrm{~atm}\) and \(273 \mathrm{~K}\). What is the mol. wt. of gas ? (a) 64 (b) 80 (c) 96 (d) None of these

Short Answer

Expert verified
The molecular weight of the gas is 32 g/mol.

Step by step solution

01

Use the Ideal Gas Law

The Ideal Gas Law equation is given by PV=nRT, where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature. We can solve for n, the number of moles, since P, V, T, and R are known.
02

Calculate the number of moles

Substitute the values into the Ideal Gas Law equation: 1 atm for P, 5.6 L for V, 0.0821 L atm/mol K for R (the standard gas constant), and 273 K for T. Calculate n using the formula n = PV/(RT).
03

Calculate the molecular weight (Molar Mass)

The molecular weight is found by dividing the mass of the gas by the number of moles. Use the mass given (20g) and the number of moles calculated from step 2 to find the molecular weight.

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Key Concepts

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

Molar Mass Calculation
Understanding how to calculate the molar mass of a substance is a fundamental skill in chemistry. Molar mass is defined as the mass of one mole of a substance and is expressed in grams per mole (g/mol). It is essentially the sum of the atomic masses of all the atoms in the molecule. To calculate the molar mass in the context of an ideal gas, you can use the Ideal Gas Law, which is represented as PV=nRT, where n is the number of moles.

For instance, when given the mass of a gas and its volume, pressure, and temperature conditions, the number of moles can be calculated using the Ideal Gas Law. Subsequently, the molar mass is found by dividing the mass of the gas by the number of moles. This step is crucial in determining the molecular formula of an unknown gas in lab experiments or on standardized tests such as the JEE.
Gas Constant
The gas constant, symbolized as R, is an important part of the Ideal Gas Law and other equations in thermodynamics. It is a proportionality constant that relates the energy scale to the temperature scale, allowing us to use temperature in calculations involving energy. In the context of the Ideal Gas Law, R bridges the relationship between pressure (P), volume (V), number of moles (n), and temperature (T).

The value of R depends on the units used for pressure, volume, and temperature. For example, in the case of the Ideal Gas Law, when using standard atmospheric pressure (atm) and volume in liters (L), the value of R is 0.0821 L atm/mol K. Knowing the correct value of R is pivotal to accurately calculate the number of moles or other unknown variables in gas law problems.
Molecular Weight Determination
Molecular weight, or molecular mass, refers to the mass of a molecule and is measured in atomic mass units (amu). In chemistry, it is typically used interchangeably with the term molar mass, though there’s a slight difference. The molecular weight is the mass of one molecule, whereas the molar mass is the weight of Avogadro's number of molecules (one mole).

Determining the molecular weight is essential when dealing with reactions and stoichiometry because it helps in calculating the amount of reactants and products. In the example scenario, the molecular weight determination is linked to an ideal gas, which follows the behavior outlined by the Ideal Gas Law. By combining the mass of the gas with the moles calculated using the Ideal Gas Law, one can deduce the molecular weight of the gas. This is especially useful when identifying unknown compounds or when the molecular formula is not given.
JEE Physical Chemistry
The Joint Entrance Examination (JEE) is a highly competitive exam for engineering aspirants in India. Physical Chemistry is one of its core sections and demands a profound understanding of concepts like the Ideal Gas Law, acid-base reactions, electrochemistry, thermodynamics, and kinetics. Problems related to the Ideal Gas Law, such as calculating molar mass or molecular weight determination, are common in the JEE Physical Chemistry paper.

Competency in solving these problems requires an understanding of the equations, constants like the gas constant, and a methodical approach to applying these concepts. Familiarity with practical examples and consistent practice with a variety of problems significantly improves a student's ability to tackle the Physical Chemistry portion of the JEE. In context, solving problems that require the use of the Ideal Gas Law to calculate molar masses contributes to the foundational skills tested in JEE Physical Chemistry.

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Most popular questions from this chapter

In an organic compound of molar mass \(108 \mathrm{gm} \mathrm{mol}^{-1} \mathrm{C}, \mathrm{H}\) and \(\mathrm{N}\) atoms are present in \(9: 1:\) 3.5 by weight. Molecular formula can be : (a) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{2}\) (b) \(\mathrm{C}_{7} \mathrm{H}_{10} \mathrm{~N}\) (c) \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{~N}_{3}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{18} \mathrm{~N}_{3}\)

A sample of calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) has the following percentage composition : \(\mathrm{Ca}=40 \%\), \(\mathrm{C}=12 \%, \mathrm{O}=48 \%\). If the law of constant proportions is true, then the weight of calcium in \(4 \mathrm{~g}\) of a sample of calcium carbonate obtained from another source will be : (a) \(0.016 \mathrm{~g}\) (b) \(0.16 \mathrm{~g}\) (c) \(1.6 \mathrm{~g}\) (d) \(16 \mathrm{~g}\)

What is the molartiy of \(\mathrm{SO}_{4}^{2-}\) ion in aqueous solution that contain \(34.2 \mathrm{ppm}\) of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) ? (Assume complete dissociation and density of solution \(1 \mathrm{~g} / \mathrm{mL}\) ) (a) \(3 \times 10^{-4} M\) (b) \(2 \times 10^{-4} M\) (c) \(10^{-4} M\) (d) None of these

The relation between molarity \((M)\) and molality \((m)\) is given by : \(\left(\rho=\right.\) density of solution \((\mathrm{mg} / \mathrm{mL}), M_{1}=\) molecular weight of solute) (a) \(m=\frac{1000 M}{1000 \rho-M_{1}}\) (b) \(m=\frac{1000 \rho M}{1000 \rho-M M_{1}}\) (c) \(m=\frac{1000 \mathrm{MM}}{1000 \rho-M M_{1}}\) (d) \(m=\frac{1000 M}{1000 \rho-M M_{1}}\)

Wood's metal contains \(50.0 \%\) bismuth, \(25.0 \%\) lead, \(12.5 \%\) tin and \(12.5 \%\) cadmium by weight. What is the mole fraction of tin ? (Atomic weights : \(\mathrm{Bi}=209, \mathrm{~Pb}=207, \mathrm{Sn}=119, \mathrm{Cd}=112\) ) (a) \(0.202\) (b) \(0.158\) (c) \(0.176\) (d) \(0.221\)

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