Problem 11
Question
The correct relationship between molecular mass and vapour density is (a) V.D. \(=2 \mathrm{M}\) (b) V.D. \(=\frac{\mathrm{M}}{2}\) (c) \(\mathrm{M}=(\mathrm{V} . \mathrm{D})^{\frac{1}{2}}\) (d) \(\mathrm{VD}=\mathrm{M}^{2}\)
Step-by-Step Solution
Verified Answer
The correct relationship is option (b): V.D. \(=\frac{\mathrm{M}}{2}\).
1Step 1: Understanding Vapour Density
Vapour density (V.D.) of a gas is defined as the mass of a certain volume of the gas divided by the mass of the same volume of hydrogen. This can be expressed as \( \text{V.D.} = \frac{\text{Molar mass of gas}}{2} \) because the molar mass of hydrogen (H\(_2\)) is 2 grams per mole.
2Step 2: Deriving the Relationship
As stated, \( \text{V.D.} = \frac{\text{Molar mass of gas}}{2} \). Rearranging this equation to express molar mass (M) in terms of vapour density gives \( \text{M} = 2 \times \text{V.D.} \).
3Step 3: Matching with the Options
Check the derived formula \( \text{M} = 2 \times \text{V.D.} \) against the provided options. It matches with option (a), where \( \text{V.D.} = \frac{\text{M}}{2} \). Therefore, the correct answer is option (b), \( \text{V.D.} = \frac{\text{M}}{2} \).
Key Concepts
Molecular MassMolar MassGas Laws
Molecular Mass
Molecular mass, also known as molecular weight, is a crucial concept in chemistry.
It is the sum of the atomic masses of all the atoms that make up a molecule. These atomic masses are usually measured in atomic mass units (amu).
For instance, the molecular mass of water (H to H-atom = 1 amu O-atom = 16 amu H Therefore, molecular mass = 2(1) + 16 = 18 amu.
Understanding molecular mass helps in many areas of chemistry. It allows us to calculate the amount of one substance that reacts with a given amount of another substance.
It is the sum of the atomic masses of all the atoms that make up a molecule. These atomic masses are usually measured in atomic mass units (amu).
For instance, the molecular mass of water (H to H-atom = 1 amu O-atom = 16 amu H Therefore, molecular mass = 2(1) + 16 = 18 amu.
Understanding molecular mass helps in many areas of chemistry. It allows us to calculate the amount of one substance that reacts with a given amount of another substance.
- This is especially important in stoichiometry, where balanced chemical equations are used.
- Molecular mass is also central in determining concentrations, as it helps convert between masses and moles.
Molar Mass
Molar mass is an extension of molecular mass. While molecular mass is the sum of atomic masses, molar mass is the mass of one mole of a substance.
It is usually expressed in grams per mole (g/mol).
To convert molecular mass in amu to molar mass, we follow a simple rule: one mole of a substance has a mass equal to its molecular mass expressed in grams.
For example, if the molecular mass of carbon dioxide (COO = 32 amu, yielding a molar mass: 44 g/mol.
Molar mass allows for the exciting possibility of predicting how different substances will interact and how much product is expected from a given reactant.
It is usually expressed in grams per mole (g/mol).
To convert molecular mass in amu to molar mass, we follow a simple rule: one mole of a substance has a mass equal to its molecular mass expressed in grams.
For example, if the molecular mass of carbon dioxide (COO = 32 amu, yielding a molar mass: 44 g/mol.
- Molar mass is crucial in laboratory settings, helping chemists accurately measure substances.
- It acts as a bridge to connect the micro world of atoms to the macro world of gram quantities in labs.
Molar mass allows for the exciting possibility of predicting how different substances will interact and how much product is expected from a given reactant.
Gas Laws
Gas laws describe the behavior of gases under various conditions of pressure, temperature, and volume.
These laws are vital for predicting and understanding how gases will respond when subjected to changes in these conditions.
The combination of these laws is represented in the Ideal Gas Law, expressed as \( PV = nRT \), where:- \( P \) is the pressure,- \( V \) is the volume,- \( n \) is the number of moles,- \( R \) is the universal gas constant,- \( T \) is the temperature in Kelvin.
Understanding gas laws is essential for many practical applications. These include predicting weather patterns and designing equipment like scuba tanks.
They also provide a foundational understanding essential for more complex concepts in thermodynamics and kinetics.
These laws are vital for predicting and understanding how gases will respond when subjected to changes in these conditions.
- Boyle’s Law shows that the pressure of a gas is inversely proportional to its volume at a constant temperature.
- Charles’s Law indicates that the volume of a gas is directly proportional to its temperature at a constant pressure.
- Avogadro’s Law states equal volumes of gases contain an equal number of molecules at the same temperature and pressure.
The combination of these laws is represented in the Ideal Gas Law, expressed as \( PV = nRT \), where:- \( P \) is the pressure,- \( V \) is the volume,- \( n \) is the number of moles,- \( R \) is the universal gas constant,- \( T \) is the temperature in Kelvin.
Understanding gas laws is essential for many practical applications. These include predicting weather patterns and designing equipment like scuba tanks.
They also provide a foundational understanding essential for more complex concepts in thermodynamics and kinetics.
Other exercises in this chapter
Problem 7
The number of oxygen atoms in \(4.4 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) is (a) \(6 \times 10^{23}\) (b) \(11 \times 10^{23}\) (c) \(12 \times 10^{23}\) (d) \(1
View solution Problem 9
The number of moles of \(\mathrm{KCl}\) in \(1000 \mathrm{~mL}\) of 3 molar solution is (a) 2 (b) 3 (c) 4 (d) 6
View solution Problem 12
A bivalent metal has an equivalent mass of 32, the molecular mass of metal nitrate is? (a) 168 (b) 182 (c) 184 (d) 188
View solution Problem 13
How many moles of potassium chlorate should be decomposed completely to obtain \(67.2\) litres of oxygen at STP? (a) 1 (b) 2 (c) 3 (d) 4
View solution