States Of Matter

The density of water never has absolute value because it varies with temperature. The density of water vapour at boiling point of water is

Step 1: Recall the stability trend of alkali metal carbonates.
Thermal stability of carbonates increases down the group.
Step 2: Analyze lithium carbonate.
Lithium carbonate is unstable due to the small size and high polarizing power of .
Step 3: Write the decomposition reaction.

Step 4: Conclusion.
Only lithium carbonate decomposes on heating to release CO .

Step 1: Write the relation between and EMF.

Step 2: Identify the values.
For the given reaction, number of electrons transferred .
Faraday constant .
EMF .
Step 3: Substitute the values.


Step 4: Convert joules to kilojoules.

Step 5: Conclusion.
The value of Gibbs free energy change is –11.4 kJ.

Step 1: Understand polarisation.
Polarisation of an atom or molecule depends on the size of its electron cloud and the ease with which it can be distorted by an external electric field.
Step 2: Compare noble gases.
Among noble gases, polarizability increases down the group due to increasing atomic size and number of electrons.
Step 3: Analyze the options.
Helium and neon are very small atoms and are least polarizable. Argon is larger than neon but smaller than krypton. Krypton has the largest atomic size among the given options.
Step 4: Conclusion.
Krypton is most easily polarised by water among the given noble gases.

Step 1: Define isochoric process.
An isochoric process is a thermodynamic process in which the volume of the gas remains constant throughout the change.
Step 2: Apply the gas law.
At constant volume, pressure is directly proportional to temperature according to Gay-Lussac’s law:

Step 3: Analyze the options.
(A) This represents Charles’ law at constant pressure.
(B) This correctly represents an isochoric process, where pressure varies with temperature at constant volume.
(C) This represents Boyle’s law at constant temperature.
(D) This also relates to Boyle’s law.
Step 4: Conclusion.
An isochor is plotted between temperature and pressure at constant volume.

Step 1: Understanding the process of liquefaction.
Liquefaction of gases depends on their ability to form intermolecular forces, such as Van der Waals forces. Oxygen has relatively weak intermolecular forces compared to gases like ammonia and carbon dioxide, which are easier to liquefy due to stronger intermolecular interactions.
Step 2: Analyzing the options.
(A) Ammonia: Incorrect. Ammonia has strong hydrogen bonding, which makes it easier to liquefy.
(B) Carbon dioxide: Incorrect. Carbon dioxide has relatively strong intermolecular forces and is easily liquefied.
(C) Oxygen: Correct — Oxygen has weak intermolecular forces, making it harder to liquefy under normal conditions.
(D) Chlorine: Incorrect. Chlorine can be liquefied at low temperatures due to its relatively stronger intermolecular forces.
Step 3: Conclusion.
The correct answer is (C) Oxygen, as it is the gas most difficult to liquefy among the options.

Step 1: Identify the applicable law.
At constant pressure, Charles’ law applies: Step 2: Write the given values.
At STP, The volume becomes double: Step 3: Calculate the final temperature.
Step 4: Convert into degree Celsius.

A Body-Centered Cubic (BCC) structure is a type of crystal lattice arrangement that consists of atoms positioned at specific points within a cubic unit cell.
Step 1: Lattice Points in a BCC Structure. In a BCC structure, atoms are arranged as follows: - There are 8 atoms at the corners of the cube. - There is 1 atom at the center of the cube.
Step 2: Contribution of Each Lattice Point. - The atoms located at the corners of the cube are shared by 8 adjacent unit cells, so each corner atom contributes of an atom to the unit cell. - The center atom is entirely contained within the unit cell and contributes 1 atom to the unit cell.
Step 3: Calculate the Total Number of Lattice Points. To calculate the total number of lattice points in the unit cell:
Conclusion: The Body-Centered Cubic (BCC) structure contains lattice points per unit cell. Thus, the correct answer is .

We use the ideal gas equation for STP conditions where: - 1 mole of any ideal gas occupies 22.4 L at STP. Now, calculate the number of moles in 1.12 L of the gas: The molar mass of the gas is: Thus, the molar mass of the gas is 50 g/mol. However, this calculation assumes standard conditions.