CBSE Class 11 Chemistry Chapter 2: Structure of Atom NCERT Solutions

Solution:

1. To find the total number of electrons in one mole of methane (CH$_4$), we first determine the number of electrons in a single molecule.

The atomic number of Carbon (C) is 6, so it has 6 electrons.

The atomic number of Hydrogen (H) is 1, so it has 1 electron.

In one molecule of methane (CH$_4$), the total number of electrons is:

$$Total\backslash \; electrons\backslash \; in\backslash \; CH\_4\; =\; (1\; \backslash times\; \backslash text\{electrons\backslash \; in\backslash \; C\})\; +\; (4\; \backslash times\; \backslash text\{electrons\backslash \; in\backslash \; H\})$$

$$Total\backslash \; electrons\backslash \; in\backslash \; CH\_4\; =\; (1\; \backslash times\; 6)\; +\; (4\; \backslash times\; 1)\; =\; 6\; +\; 4\; =\; 10\backslash \; electrons$$

One mole of any substance contains Avogadro's number of particles, which is $6.022\; \backslash times\; 10^\{23\}$.

Therefore, one mole of methane contains $6.022\; \backslash times\; 10^\{23\}$ molecules.

The total number of electrons in one mole of methane is:

$$Total\backslash \; electrons\backslash \; in\backslash \; 1\backslash \; mole\backslash \; of\backslash \; CH\_4\; =\; (Number\backslash \; of\backslash \; electrons\backslash \; per\backslash \; molecule)\; \backslash times\; (Avogadro\text{'}s\backslash \; number)$$

$$Total\backslash \; electrons\backslash \; in\backslash \; 1\backslash \; mole\backslash \; of\backslash \; CH\_4\; =\; 10\; \backslash times\; (6.022\; \backslash times\; 10^\{23\})\; =\; 6.022\; \backslash times\; 10^\{24\}\backslash \; electrons$$

2. (a) Total number of neutrons in 7 mg of $^{14}$C:

The isotope $^{14}$C has an atomic number of 6 and a mass number of 14.

The number of neutrons in one atom of $^{14}$C is (Mass number - Atomic number) = $14\; -\; 6\; =\; 8$ neutrons.

One mole of $^{14}$C atoms weighs 14 grams and contains Avogadro's number ($6.022\; \backslash times\; 10^\{23\}$) of atoms.

So, 14 g of $^{14}$C contains $6.022\; \backslash times\; 10^\{23\}$ atoms, each with 8 neutrons.

Total neutrons in 14 g of $^{14}$C = $8\; \backslash times\; (6.022\; \backslash times\; 10^\{23\})\; =\; 4.8176\; \backslash times\; 10^\{24\}$ neutrons.

We need to find the number of neutrons in 7 mg of $^{14}$C. First, convert 7 mg to grams: $7\; \backslash text\{\; mg\}\; =\; 7\; \backslash times\; 10^\{-3\}\; \backslash text\{\; g\}$.

Using a proportion:

$$Number\backslash \; of\backslash \; neutrons\backslash \; in\backslash \; 7\backslash \; mg\; =\; \backslash frac\{\backslash text\{Total\backslash \; neutrons\backslash \; in\backslash \; 14\backslash \; g\}\}\{14\backslash \; g\}\; \backslash times\; 7\; \backslash times\; 10^\{-3\}\backslash \; g$$

$$Number\backslash \; of\backslash \; neutrons\backslash \; in\backslash \; 7\backslash \; mg\; =\; \backslash frac\{4.8176\; \backslash times\; 10^\{24\}\}\{14\}\; \backslash times\; (7\; \backslash times\; 10^\{-3\})$$

$$Number\backslash \; of\backslash \; neutrons\backslash \; in\backslash \; 7\backslash \; mg\; =\; (0.3441\; \backslash times\; 10^\{24\})\; \backslash times\; (7\; \backslash times\; 10^\{-3\})\; =\; 2.4087\; \backslash times\; 10^\{21\}$$

(b) Total mass of neutrons in 7 mg of $^{14}$C:

The mass of one neutron is given as $1.675\; \backslash times\; 10^\{-27\}$ kg.

Total mass of neutrons = (Number of neutrons) $\backslash times$ (Mass of one neutron)

$$Total\backslash \; mass\backslash \; of\backslash \; neutrons\; =\; (2.4087\; \backslash times\; 10^\{21\})\; \backslash times\; (1.675\; \backslash times\; 10^\{-27\}\; \backslash text\{\; kg\})$$

$$Total\backslash \; mass\backslash \; of\backslash \; neutrons\; =\; 4.034\; \backslash times\; 10^\{-6\}\; \backslash text\{\; kg\}$$

3. (a) Total number of protons in 34 mg of NH$_3$:

First, find the number of protons in one molecule of ammonia (NH$_3$).

Nitrogen (N) has an atomic number of 7, so it has 7 protons.

Hydrogen (H) has an atomic number of 1, so it has 1 proton.

In one molecule of NH$_3$, the total number of protons is:

$$Total\backslash \; protons\backslash \; in\backslash \; NH\_3\; =\; (1\; \backslash times\; \backslash text\{protons\backslash \; in\backslash \; N\})\; +\; (3\; \backslash times\; \backslash text\{protons\backslash \; in\backslash \; H\})$$

$$Total\backslash \; protons\backslash \; in\backslash \; NH\_3\; =\; (1\; \backslash times\; 7)\; +\; (3\; \backslash times\; 1)\; =\; 7\; +\; 3\; =\; 10\backslash \; protons$$

Now, we need to find the number of NH$_3$ molecules in 34 mg.

The molar mass of NH$_3$ is (Atomic mass of N + 3 $\backslash times$ Atomic mass of H) = $(14.007\; +\; 3\; \backslash times\; 1.008)\; \backslash approx\; 17.031$ g/mol.

So, 17.031 g of NH$_3$ contains $6.022\; \backslash times\; 10^\{23\}$ molecules.

We have 34 mg of NH$_3$, which is $34\; \backslash times\; 10^\{-3\}$ g.

Number of moles of NH$_3$ = $\backslash frac\{\backslash text\{Mass\}\}\{\backslash text\{Molar\backslash \; mass\}\}\; =\; \backslash frac\{34\; \backslash times\; 10^\{-3\}\backslash \; g\}\{17.031\backslash \; g/mol\}\; \backslash approx\; 2\; \backslash times\; 10^\{-3\}$ moles.

Number of molecules in 34 mg of NH$_3$ = (Number of moles) $\backslash times$ (Avogadro's number)

$$Number\backslash \; of\backslash \; molecules\; =\; (2\; \backslash times\; 10^\{-3\})\; \backslash times\; (6.022\; \backslash times\; 10^\{23\})\; =\; 1.2044\; \backslash times\; 10^\{21\}$$

Total number of protons in 34 mg of NH$_3$ = (Number of molecules) $\backslash times$ (Protons per molecule)

$$Total\backslash \; protons\; =\; (1.2044\; \backslash times\; 10^\{21\})\; \backslash times\; 10\; =\; 1.2044\; \backslash times\; 10^\{22\}$$

(b) Total mass of protons in 34 mg of NH$_3$:

The mass of one proton is approximately $1.67493\; \backslash times\; 10^\{-27\}$ kg.

Total mass of protons = (Total number of protons) $\backslash times$ (Mass of one proton)

$$Total\backslash \; mass\backslash \; of\backslash \; protons\; =\; (1.2044\; \backslash times\; 10^\{22\})\; \backslash times\; (1.67493\; \backslash times\; 10^\{-27\}\; \backslash text\{\; kg\})$$

$$Total\backslash \; mass\backslash \; of\backslash \; protons\; =\; 2.017\; \backslash times\; 10^\{-5\}\; \backslash text\{\; kg\}$$

Will the answer change if the temperature and pressure are changed?

No, the answer will not change if the temperature and pressure are changed. The number of protons and their mass are intrinsic properties of the atoms and molecules. Changes in temperature and pressure affect the volume and density of a gas, but not the number of protons within a given mass of the substance.