CBSE Class 9 Science Exemplar Chapter 3: Atoms and Molecules NCERT Solutions

To determine the correct representation of 360 g of water, we need to calculate the number of moles and the number of molecules.

The molar mass of water (H_2O) is calculated as follows:

Atomic mass of Hydrogen (H) = 1 g/mol

Atomic mass of Oxygen (O) = 16 g/mol

Molar mass of H_2O = (2 \times 1) + 16 = 18 g/mol.

Now, we calculate the number of moles in 360 g of water:

Number of moles = \frac{\text{Mass of substance}}{\text{Molar mass of substance}}

Number of moles = \frac{360 \text{ g}}{18 \text{ g/mol}} = 20 \text{ moles}.

This matches option (ii).

Next, we calculate the number of molecules using Avogadro's number (6.022 \times 10^{23} molecules/mol):

Number of molecules = Number of moles \times Avogadro's number

Number of molecules = 20 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mol}

Number of molecules = 1.2044 \times 10^{25} molecules.

This matches option (iv).

Therefore, the correct options are (ii) and (iv).

Answer: (d) (ii) and (iv)

Let's analyze each statement:

(a) Atoms are not able to exist independently: This is often true for many atoms, especially non-metals, which tend to form bonds to achieve stability. Noble gases are an exception.

(b) Atoms are the basic units from which molecules and ions are formed: This is a fundamental concept in chemistry. Atoms combine to form molecules, and they can gain or lose electrons to form ions.

(c) Atoms are always neutral in nature: An atom consists of a neutral nucleus (protons and neutrons) and electrons orbiting it. The number of protons (positive charge) equals the number of electrons (negative charge), resulting in a net neutral charge.

(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch: While atoms do aggregate to form matter, individual atoms are extremely small and cannot be perceived by our senses (see, feel, or touch) directly. We perceive the bulk matter formed by a vast number of atoms.

The statement that is not true is that we can see, feel, or touch the matter formed by atoms directly with our naked eyes.

Answer: (d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch

The chemical symbol for the element nitrogen is 'N'. However, nitrogen gas, as it exists in its elemental form under standard conditions, is a diatomic molecule. This means that two nitrogen atoms are bonded together to form a stable molecule.

Therefore, when referring to nitrogen gas, the correct chemical symbol is N_2.

Option (a) 'Ni' is the symbol for Nickel.

Option (c) 'N+' represents a nitrogen ion with a positive charge.

Option (d) 'N' represents a single nitrogen atom.

Answer: (b) N_2

The chemical symbol for the element sodium is 'Na'. This symbol is derived from the Latin name for sodium, which is 'Natrium'. Many chemical symbols are based on the Latin or Greek names of the elements.

Answer: (d) Na

To determine which option weighs the highest, we need to calculate the mass of each sample using the formula: Mass = Number of moles \times Molar mass.

First, let's find the molar masses of the substances involved:

Molar mass of Sucrose (C_{12} H_{22} O_{11}) = (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 g/mol.

Molar mass of Carbon Dioxide (CO_2) = 12 + (2 \times 16) = 12 + 32 = 44 g/mol.

Molar mass of Calcium Carbonate (CaCO_3) = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100 g/mol.

Molar mass of Water (H_2O) = (2 \times 1) + 16 = 18 g/mol.

Now, let's calculate the mass for each option:

(a) Mass of 0.2 mole of sucrose = 0.2 \text{ mol} \times 342 \text{ g/mol} = 68.4 \text{ g}.

(b) Mass of 2 moles of CO_2 = 2 \text{ mol} \times 44 \text{ g/mol} = 88 \text{ g}.

(c) Mass of 2 moles of CaCO_3 = 2 \text{ mol} \times 100 \text{ g/mol} = 200 \text{ g}.

(d) Mass of 10 moles of H_2O = 10 \text{ mol} \times 18 \text{ g/mol} = 180 \text{ g}.

Comparing the masses, 200 g is the highest.

Answer: (c) 2 moles of CaCO_3

To find the maximum number of atoms, we need to calculate the number of moles for each substance and then multiply by the number of atoms per molecule and Avogadro's number.

First, calculate the molar masses:

Molar mass of H_2O = 18 g/mol.

Molar mass of O_2 = 32 g/mol.

Molar mass of CO_2 = 44 g/mol.

Molar mass of CH_4 = 12 + (4 \times 1) = 16 g/mol.

Now, calculate the number of moles for 18g of each substance:

(a) Moles of H_2O = \frac{18 \text{ g}}{18 \text{ g/mol}} = 1 \text{ mole}. Number of atoms = 1 mole \times 3 atoms/molecule \times 6.022 \times 10^{23} molecules/mol = 3 \times 6.022 \times 10^{23} atoms.

(b) Moles of O_2 = \frac{18 \text{ g}}{32 \text{ g/mol}} = 0.5625 \text{ mole}. Number of atoms = 0.5625 mole \times 2 atoms/molecule \times 6.022 \times 10^{23} molecules/mol = 1.125 \times 6.022 \times 10^{23} atoms.

(c) Moles of CO_2 = \frac{18 \text{ g}}{44 \text{ g/mol}} \approx 0.409 \text{ mole}. Number of atoms = 0.409 mole \times 3 atoms/molecule \times 6.022 \times 10^{23} molecules/mol = 1.227 \times 6.022 \times 10^{23} atoms.

(d) Moles of CH_4 = \frac{18 \text{ g}}{16 \text{ g/mol}} = 1.125 \text{ mole}. Number of atoms = 1.125 mole \times 5 atoms/molecule \times 6.022 \times 10^{23} molecules/mol = 5.625 \times 6.022 \times 10^{23} atoms.

Comparing the number of atoms, 5.625 \times 6.022 \times 10^{23} is the largest.

Answer: (d) 18g of CH_4