Question

The vapour density of gas \(A\) is four times that of \(B\). If the molecular mass of \(\mathrm{B}\) is \(\mathrm{M}\) then the molecular mass of \(A\) is (1) \(\mathrm{M}\) (2) \(2 \mathrm{M}\) (3) \(\mathrm{M} / 4\) (4) \(4 \mathrm{M}\)

Short answer

The molecular mass of \(A\) is \(4M\).

Step-by-step solution

Understand the concept of vapour density

Vapour density of a gas is the ratio of the mass of a certain volume of the gas to the mass of the same volume of hydrogen at the same temperature and pressure. It is given by the formula: \[ \text{Vapour Density (VD)} = \frac{\text{Molecular Mass}}{2} \]

Express the given relationship

We know from the problem that the vapour density of gas \(A\) is four times that of gas \(B\). This can be written as: \[ \text{VD}_A = 4 \times \text{VD}_B \]

Substitute the vapour density formula

Replace the vapour densities with their respective molecular mass equations: \[ \frac{M_A}{2} = 4 \times \frac{M_B}{2} \]

Simplify the equation

Solving the equation, we get: \[ \frac{M_A}{2} = 4 \times \frac{M}{2} \] \[ M_A = 4M \]

Conclusion

Thus, the molecular mass of gas \(A\) is four times the molecular mass of gas \(B\). So the correct option is: (4) \(4M\)

Key concepts

molecular mass

Molecular mass, or molecular weight, is the sum of the atomic masses of all the atoms in a molecule. It is usually expressed in atomic mass units (amu) or Daltons (Da). To find the molecular mass, you add up the atomic masses of all the elements present in the molecule. For example, the molecular mass of water (H₂O) is calculated as follows: the atomic mass of hydrogen (H) is approximately 1 amu, and that of oxygen (O) is roughly 16 amu. Therefore, the molecular mass of H₂O is (2 × 1 amu) + 16 amu = 18 amu. Understanding molecular mass is crucial, as it helps us determine the amount of substance present in chemical reactions and stoichiometric calculations.

gas laws

Gas laws describe the behavior of gases under different conditions of temperature, pressure, and volume. One of the most fundamental gas laws is the Ideal Gas Law, which is expressed as PV = nRT , where P represents pressure, V stands for volume, n is the number of moles of the gas, R is the gas constant, and T is the temperature in Kelvin. Another important law is Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure when temperature is held constant: P₁V₁ = P₂V₂ . These laws are essential for understanding how gases will react and change under varying conditions, which is especially important in fields such as chemistry, physics, and engineering.

stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships of the elements and compounds involved in chemical reactions. It allows chemists to predict the amounts of reactants and products involved in a given reaction. To perform stoichiometric calculations, you need a balanced chemical equation that represents the reaction. The coefficients in the balanced equation indicate the ratio of the moles of each substance involved. For example, in the reaction: 2H₂ + O₂ → 2H₂O This tells us that 2 molecules of hydrogen react with 1 molecule of oxygen to produce 2 molecules of water. Using these ratios, you can calculate the amount of product formed from a given amount of reactant, which is critical in industrial processes, laboratory experiments, and many real-world applications.

chemical calculations

Chemical calculations are the heart of quantitative chemistry. These calculations often involve converting units, using molar masses, and employing stoichiometric relationships. One common calculation is converting between mass and moles using the formula: mass (g) = moles × molar mass (g/mol) . Another key type of calculation is determining the yield of a reaction, which tells us the efficiency of a chemical reaction. The percent yield can be calculated using: percent yield = (actual yield / theoretical yield) × 100% . These calculations enable chemists to measure reactants, analyze products, and optimize chemical processes. Precise chemical calculations ensure consistent and reliable results in both academic and industrial chemistry.