NCERT Solutions: Chapter 1: Some Basic Concepts of Chemistry.

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1.1 Calculate the molar mass of the following:

• (i) $H_2O$: $(2 \times 1.008) + 16.00 = \mathbf{18.016 \, \text{g/mol}}$

• (ii) $CO_2$: $12.01 + (2 \times 16.00) = \mathbf{44.01 \, \text{g/mol}}$

• (iii) $CH_4$: $12.01 + (4 \times 1.008) = \mathbf{16.042 \, \text{g/mol}}$

1.2 Calculate the mass per cent of different elements present in sodium sulphate ($Na_2SO_4$).

Step 1: Find Molar Mass of $Na_2SO_4$

$(2 \times 23.0) + 32.06 + (4 \times 16.00) = 142.06 \, \text{g/mol}$

Step 2: Calculate Mass % ($\dfrac{\text{Mass of element}}{\text{Molar mass}} \times 100$)

• Sodium (Na): $\dfrac{46.0}{142.06} \times 100 = \mathbf{32.38\%}$

• Sulphur (S): $\dfrac{32.06}{142.06} \times 100 = \mathbf{22.57\%}$

• Oxygen (O): $\dfrac{64.0}{142.06} \times 100 = \mathbf{45.05\%}$

1.3 Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Element	Mass %	Atomic Mass	Moles (Mass/At. Mass)	Molar Ratio	Simple Ratio
Fe	69.9	55.85	$69.9 / 55.85 = 1.25$	$1.25 / 1.25 = 1$	2
O	30.1	16.00	$30.1 / 16.00 = 1.88$	$1.88 / 1.25 = 1.5$	3

Empirical Formula: $\mathbf{Fe_2O_3}$

1.4 Calculate the amount of carbon dioxide produced when:

Reaction: $C(s) + O_2(g) \rightarrow CO_2(g)$

• (i) 1 mole of carbon is burnt in air: Since $O_2$ is in excess, 1 mole of $C$ gives 1 mole (44g) of $CO_2$.

• (ii) 1 mole of carbon is burnt in 16g of dioxygen: 16g $O_2 = 0.5$ mole. $O_2$ is the limiting reagent. 0.5 mole $O_2$ produces 0.5 mole (22g) of $CO_2$.

• (iii) 2 moles of carbon are burnt in 16g of dioxygen: Again, 16g $O_2 (0.5 \text{ mole})$ is the limiting reagent. It can only react with 0.5 mole of $C$ to produce 0.5 mole (22g) of $CO_2$.

1.5 Calculate the mass of sodium acetate ($CH_3COONa$) required to make 500 mL of 0.375 molar aqueous solution.

Step 1: Use Molarity formula ($M = \dfrac{n}{V}$)

$0.375 = \dfrac{n}{0.500 \, \text{L}} \implies n = 0.375 \times 0.500 = 0.1875 \, \text{moles}$

Step 2: Convert moles to mass (Molar mass = 82.0245 g/mol)

$\text{Mass} = 0.1875 \, \text{mol} \times 82.0245 \, \text{g/mol} = \mathbf{15.38 \, \text{g}}$

1.6 Calculate the concentration of nitric acid ($HNO_3$) in moles per litre ($M$) in a sample with density 1.41 g/mL and 69% mass per cent.

Step 1: Find mass of 1L solution

$\text{Mass} = \text{Volume} \times \text{Density} = 1000 \, \text{mL} \times 1.41 \, \text{g/mL} = 1410 \, \text{g}$

Step 2: Find mass of $HNO_3$ in that 1L

$69\% \text{ of } 1410 \, \text{g} = 0.69 \times 1410 = 972.9 \, \text{g}$

Step 3: Calculate Molarity (Molar mass of $HNO_3 = 63 \, \text{g/mol}$)

$M = \dfrac{972.9 \, \text{g} / 63 \, \text{g/mol}}{1 \, \text{L}} = \mathbf{15.44 \, \text{M}}$

1.6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL⁻¹ and the mass per cent of nitric acid in it being 69%.

Step 1: Find the mass of 1 L of the solution

• $\text{Density} = 1.41\text{ g/mL}$

• $\text{Mass of 1000 mL solution} = 1000 \times 1.41 = 1410\text{ g}$

Step 2: Find the mass of actual } HNO_3 \text{ in the solution

• The solution is $69\%$ $HNO_3$ by mass.

• $\text{Mass of } HNO_3 = \dfrac{69}{100} \times 1410\text{ g} = 972.9\text{ g}$

Step 3: Calculate Molarity (Moles per litre)

• $\text{Molar mass of } HNO_3 = 1 + 14 + (3 \times 16) = 63\text{ g/mol}$

• $\text{Moles of } HNO_3 = \dfrac{972.9\text{ g}}{63\text{ g/mol}} = 15.44\text{ mol}$

• Concentration = 15.44 M

1.7 How much copper can be obtained from 100 g of copper sulphate ($CuSO_4$)?

Step 1: Calculate the molar mass of $CuSO_4$

• $\text{Atomic mass of Cu} = 63.5\text{ u}$, $\text{S} = 32\text{ u}$, $\text{O} = 16\text{ u}$

• $\text{Molar mass} = 63.5 + 32 + (4 \times 16) = 159.5\text{ g/mol}$

Step 2: Use the ratio to find the mass of Copper

• $159.5\text{ g}$ of $CuSO_4$ contains $63.5\text{ g}$ of Copper.

• $\text{Copper in 100 g } CuSO_4 = \left(\dfrac{63.5}{159.5}\right) \times 100 = \mathbf{39.81\text{ g}}$

1.8 Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.

(Note: Molar mass of this oxide is 159.69 g/mol)

Step 1: Find the Empirical Formula

• $\text{Moles of Fe} = \dfrac{69.9}{55.85} = 1.25$

• $\text{Moles of O} = \dfrac{30.1}{16.00} = 1.88$

• $\text{Simple molar ratio (Fe:O)} = 1 : 1.5$, which simplifies to 2 : 3.

• $\text{Empirical Formula} = Fe_2O_3$

Step 2: Determine Molecular Formula

• $\text{Empirical Formula Mass} = (2 \times 55.85) + (3 \times 16) = 159.7\text{ g/mol}$

• Since the molar mass ($159.69$) is approximately equal to the empirical mass, $n = 1$.

• Molecular Formula = $Fe_2O_3$

1.9 Calculate the atomic mass (average) of chlorine using the following data:

Isotope	% Natural Abundance	Molar Mass (g/mol)
$^{35}Cl$	75.77	34.9689
$^{37}Cl$	24.23	36.9659

Calculation:

$$\text{Avg. Mass} = \dfrac{(75.77 \times 34.9689) + (24.23 \times 36.9659)}{100}$$

$$\text{Avg. Mass} = \dfrac{2649.59 + 895.68}{100} = \dfrac{3545.27}{100} = \mathbf{35.4527\text{ u}}$$

1.10 In three moles of ethane ($C_2H_6$), calculate the following:

• (i) Number of moles of carbon atoms:

  1 mole of $C_2H_6$ has 2 moles of C.

  3 moles of $C_2H_6$ have $3 \times 2 = \mathbf{6\text{ moles}}$ of C atoms.

• (ii) Number of moles of hydrogen atoms:

  1 mole of $C_2H_6$ has 6 moles of H.

  3 moles of $C_2H_6$ have $3 \times 6 = \mathbf{18\text{ moles}}$ of H atoms.

• (iii) Number of molecules of ethane:

  $1\text{ mole} = 6.022 \times 10^{23}\text{ molecules}$

  $3\text{ moles} = 3 \times 6.022 \times 10^{23} = \mathbf{1.806 \times 10^{24}\text{ molecules}}$