NCERT Solutions for Class 11 Chemistry Chapter 1 – Some Basic Concepts of Chemistry

Atomic masses (approx.):
H = 1, C = 12, O = 16

(i) H₂O
H₂O = 2×H + 1×O
= 2×1 + 16
= 18 g mol⁻¹

(ii) CO₂
CO₂ = 1×C + 2×O
= 12 + 2×16
= 12 + 32
= 44 g mol⁻¹

(iii) CH₄
CH₄ = 1×C + 4×H
= 12 + 4×1
= 16
= 16 g mol⁻¹

Step 1: Take atomic masses
Na = 23, S = 32, O = 16

Step 2: Calculate molar mass of Na₂SO₄ Step 3: Formula for mass per cent (i) % of Na
Total mass of Na = 2×23 = 46 (ii) % of S
Mass of S = 32 (iii) % of O
Total mass of O = 4×16 = 64

Step 1: Assume 100 g of compound
Fe = 69.9 g
O = 30.1 g

Step 2: Calculate moles
Atomic masses: Fe = 55.85 g/mol, O = 16 g/mol Step 3: Divide by the smallest mole value
Smallest = 1.25 Ratio = Fe : O = 1 : 1.5

Step 4: Multiply by 2 to remove decimal Ratio = Fe₂ : O₃
✅ Therefore, empirical formula = Fe₂O₃

Reaction (i) 1 mole C burnt in air
In air, O₂ is available in sufficient/excess amount, so complete burning of 1 mole C occurs.

CO₂ moles = 1 mole ✅ Therefore, CO₂ = 1 mole = 44 g

(ii) 1 mole C burnt in 16 g O₂
Here O₂ is less, so first calculate moles of O₂.
(Molar mass of O₂ = 32 g/mol) From reaction, 1 mole O₂ gives 1 mole CO₂, so
0.5 mole O₂ gives 0.5 mole CO₂. ✅ Therefore, CO₂ = 0.5 mole = 22 g

(iii) 2 moles C burnt in 16 g O₂
Again, moles of O₂ = 16 ÷ 32 = 0.5 mole.
Required O₂ for 2 moles C is 2 moles, but only 0.5 mole is available.
So, O₂ is the limiting reagent. ✅ Therefore, CO₂ = 0.5 mole = 22 g

Step 1: Formula used (when volume is in mL)
Formula of molarity: Where
👉 M = molarity (mol L⁻¹)
👉 w = mass of solute (g)
👉 V = volume of solution (mL)

Step 2: Put the values
👉 M = 0.375 mol L⁻¹
👉 Molar mass = 82.0245 g mol⁻¹
👉 V = 500 mL

Now solve for w:
✅ Therefore, required mass w = 15.38 g

Step 1: Assume 100 g of solution
69% (mass per cent) means:
• In 100 g solution, HNO₃ = 69 g

Step 2: Molar mass of HNO₃
HNO₃ = 1 + 14 + (3 × 16) = 63 g mol⁻¹

Step 3: Calculate moles of HNO₃
👉 Moles of HNO₃ = mass of HNO₃ / molar mass of HNO₃
👉 Moles of HNO₃ = 69 g / 63 g mol⁻¹ = 1.095 mol

Step 4: Calculate volume of 100 g solution (using density)
Density = 1.41 g mL⁻¹
Density = mass / volume
Therefore, volume = mass / density

Volume = 100 g / 1.41 g mL⁻¹ = 70.92 mL

Step 5: Calculate molarity (M)
Molarity = moles of solute / volume of solution (in L)
70.92 mL = 0.07092 L

Molarity (M) = 1.095 mol / 0.07092 L = 15.44 mol L⁻¹
✅ Therefore, molarity of HNO₃ solution is 15.44 mol L⁻¹ (≈ 15.44 M).

Step 1: Calculate molar mass of CuSO₄
CuSO₄ = Cu + S + 4O
= 63.5 + 32 + 4×16
= 63.5 + 32 + 64
= 159.5 g mol⁻¹

Step 2: Apply unitary method
In 159.5 g CuSO₄, Cu = 63.5 g
So in 1 g CuSO₄, Cu = 63.5 / 159.5 g
Therefore in 100 g CuSO₄, Cu = (63.5 / 159.5) × 100 g ✅ Therefore, from 100 g CuSO₄, 39.81 g Cu is obtained.

Assume 100 g of compound
Then Fe = 69.9 g and O = 30.1 g

S.No.	Element	Amount in grams	Atomic mass	Moles	Simplest ratio	Integer ratio	Empirical formula part
1	Fe	69.9	55.85 (≈56)	69.9 ÷ 55.85 = 1.25	1.25 ÷ 1.25 = 1	1 × 2 = 2	Fe2
2	O	30.1	16	30.1 ÷ 16 = 1.88	1.88 ÷ 1.25 = 1.50	1.50 × 2 = 3	O3

✅ Therefore, empirical formula = Fe₂O₃

Empirical formula mass = (2 × 55.85) + (3 × 16) = 159.7 g mol^-1
n = molar mass of compound / empirical formula mass
n = 159.7 / 159.7 = 1

✅ Hence molecular formula of the given oxide:
Molecular formula = n × empirical formula
Molecular formula = 1 × Fe₂O₃
Molecular formula = Fe₂O₃

Average atomic mass = Sum of (percentage abundance × atomic mass) ÷ 100

Isotope	Natural abundance (%)	Atomic mass (u)	Product
35Cl	75.77	34.9689	75.77 × 34.9689 = 2649.59
37Cl	24.23	36.9659	24.23 × 36.9659 = 895.95

Step 2: Add
Total = 2649.59 + 895.95 = 3545.54 (approximately)

Step 3: Divide by 100
👉 Average atomic mass = 3545.27 / 100 = 35.4527 u
✅ Therefore, average atomic mass = 35.4527 u ≈ 35.45 u

Formula of ethane: C₂H₆
This means in 1 molecule:
👉 Carbon = 2 atoms
👉 Hydrogen = 6 atoms

Now for 3 moles of C₂H₆:

📌 (i) Moles of carbon atoms
In 1 mole C₂H₆, moles of carbon atoms = 2 mol
So in 3 moles = 3 × 2 = 6 mol

📌 (ii) Moles of hydrogen atoms
In 1 mole C₂H₆, moles of hydrogen atoms = 6 mol
So in 3 moles = 3 × 6 = 18 mol

📌 (iii) Number of molecules of ethane
Avogadro number (N_a) = 6.022 × 10²³ molecules per mole
So in 3 moles, molecules = 3 × 6.022 × 10²³ = 1.8066 × 10²⁴ molecules

Step 1: Calculate molar mass of sugar
Molar mass of C₁₂H₂₂O₁₁
= (12×12) + (22×1) + (11×16)
= 144 + 22 + 176
= 342 g mol^-1

Step 2: Calculate moles of sugar
Moles = mass ÷ molar mass
Moles = 20 g ÷ 342 g mol^-1 = 0.0585 mol

Step 3: Calculate molarity (molar concentration)
Given volume = 2 L
Molarity = moles ÷ volume (L)
Molarity = 0.0585 mol ÷ 2 L
Molarity = 0.02925 mol L^-1

✅ Therefore, molarity ≈ 0.0292 mol L^-1 (≈ 0.029 M)

Step 1: Molar mass of methanol
Molar mass of CH₃OH = 32 g mol⁻¹ = 0.032 kg mol⁻¹

Step 2: Calculate molarity of given pure methanol
Density (d) = 0.793 kg L⁻¹
This means mass of 1 L methanol = 0.793 kg

Moles = mass ÷ molar mass
Moles = 0.793 kg ÷ 0.032 kg mol⁻¹ = 24.78 mol
Therefore, molarity of pure methanol = 24.78 mol L⁻¹
Let this be M₁:
M₁ = 24.78 M

Step 3: Use dilution formula
M₁ × V₁ = M₂ × V₂
Where
👉 M₁ = 24.78 M (methanol)
👉 M₂ = 0.25 M (required)
👉 V₂ = 2.5 L (required)

Now find V₁:
V₁ = (M₂ × V₂) ÷ M₁

0.025 L = 25 mL
✅ Therefore, required volume of methanol = 0.025 L or 25 mL

Step 1: Formula of pressure
P = F / A
and F = mg
so P = mg / A

Given:
Mass of air at sea level = 1034 g cm⁻²
This means mass on 1 cm² area = 1034 g

Convert this mass into kg:
1034 g = 1.034 kg

Step 2: Convert area into m²
1 m = 100 cm
so 1 m² = 100 cm × 100 cm = 10,000 cm²
This means:
1 cm² = 1 / 10,000 m²

Step 3: Calculate force on 1 cm²
F = mg
m = 1.034 kg
g = 9.8 m s⁻²
F = 1.034 × 9.8 N

Step 4: Pressure in pascal
P = F / A
A = 1 cm² = 1/10000 m²
so
P = (1.034 × 9.8) / (1/10000)
P = (1.034 × 9.8) × 10000
P = 101332 Pa
P = 1.01332 × 10⁵ Pa

✅ Therefore, pressure = 1.01332 × 10⁵ Pa

1) SI unit of mass
The SI unit of mass is kilogram (kg).

2) Definition of kilogram (as per your given idea)
The mass of a special standard piece (or cylinder/block) made of platinum–iridium, kept at 0°C at Sèvres near Paris, was taken as 1 standard kilogram.
Based on this standard, all other masses were measured.

👉 micro means “one-millionth”, so 10⁻⁶.
👉 deca means “ten times”, so 10¹.
👉 mega means “one million”, so 10⁶.
👉 giga means “one billion”, so 10⁹.
👉 femto represents a very small value, so 10⁻¹⁵.
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1) Why are significant figures important?
When we measure a quantity, the reading is not perfectly exact due to instrument limitations.
So the digits we write indicate how reliable the measurement is.

2) What is included in significant figures?
👉 Certain digits: digits that can be read clearly from the instrument
👉 Estimated digit: the last digit, written by estimation

3) Examples
👉 2.35 cm → has 3 significant figures (2, 3, 5)
👉 0.00450 g → has 3 significant figures (4, 5, 0)
👉 1200 (without decimal point) → number of significant figures is ambiguous; it depends on notation
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(i) Converting 15 ppm into mass per cent

15 ppm (by mass) means 15 parts by mass of chloroform are present in 10⁶ parts by mass of water.
Mass per cent = (15 × 10^-6) × 100 = 0.0015%


(ii) Molality of CHCl₃ in water

Molar mass of CHCl₃:
C = 12.01, H = 1.008, Cl = 35.45
CHCl₃ = 12.01 + 1.008 + 3×35.45 = 119.37 g mol^-1

In 10⁶ g sample, amount of chloroform = 15 g
In 10³ g (1 kg) sample, amount of chloroform = (15 g ÷ 10⁶) × 10³ = 1.5 × 10^-2 g

Moles of chloroform in 10³ g (1 kg) sample:
Moles of chloroform = amount of chloroform ÷ molar mass of CHCl₃
= 1.5 × 10^-2 g ÷ 119.37 g mol^-1
= 1.255 × 10^-4 mol

Molality = (moles of solute) / (mass of solvent in kg)

✅ Therefore, molality = 1.26 × 10^-4 mol kg^-1

Rule for scientific notation:
Write the number so that only one non-zero digit remains to the left of decimal, then multiply by power of 10.

👉 Move decimal to the left → exponent +
👉 Move decimal to the right → exponent −

(i) 0.0048 → decimal moved 3 places right → 4.8, so 10^-3
(ii) 234,000 → decimal moved 5 places left → 2.34, so 10⁵
(iii) 8008 → decimal moved 3 places left → 8.008, so 10³
(iv) 500.0 → decimal moved 2 places left → 5.000, so 10²
(v) 6.0012 → already between 1 and 10 → 10⁰

Rules (Quick Rules)
1) Leading zeros are not significant.
2) Zeros between two non-zero digits are significant.
3) Trailing zeros after decimal are significant.
4) Trailing zeros in a whole number without decimal are generally not significant (unless specified).

Apply now:
(i) 0.0025 → only 2 and 5 count → 2
(ii) 208 → zero in between counts → 3
(iii) 5005 → both middle zeros count → 4
(iv) 126,000 → no decimal, trailing zeros not counted → 3
(v) 500.0 → decimal present, all trailing zeros count → 4
(vi) 2.0034 → zeros between non-zero digits count → 5

Rule: Keep first three significant digits, then check the next (fourth) digit.
👉 If fourth digit is 5 or more → increase third digit by 1
👉 If fourth digit is less than 5 → keep as it is

(i) 34.216
First 3 significant digits: 3, 4, 2 → 34.2
Next digit is 1 (<5) ⇒ 34.2

(ii) 10.4107
First 3 significant digits: 1, 0, 4 → 10.4
Next digit is 1 (<5) ⇒ 10.4

(iii) 0.04597
Significant digits start from 4: 4, 5, 9
Next digit is 7 (≥5) ⇒ 9 rounds up, so 0.0460
(Last zero shows 3 significant figures.)

(iv) 2808
First 3 significant digits: 2, 8, 0
Next digit is 8 (≥5) ⇒ 280 becomes 281
So 2.81 × 10³ (that is 2810)

(a) Law of Multiple Proportions
Statement:
When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in ratios of small whole numbers.

Now fix nitrogen mass at 14 g:
• (i) N = 14 g, O = 16 g
• (ii) N = 14 g, O = 32 g
O ratio = 16 : 32 = 1 : 2 (small whole numbers)

For (iii), N = 28 g, make it 14 g by dividing by 2:
• O = 32/2 = 16 g

For (iv), N = 28 g, make it 14 g by dividing by 2:
• O = 80/2 = 40 g

So with fixed N = 14 g, O masses are 16, 32, 40
Ratio = 16 : 32 : 40 = 2 : 4 : 5 (small whole numbers)
✅ Hence it follows the law of multiple proportions.

(b) Unit conversions

(i) km to mm, pm
👉 1 km = 1000 m
👉 1 m = 1000 mm ⇒ 1 km = 10³ × 10³ = 10⁶ mm
👉 1 m = 10¹² pm ⇒ 1 km = 10³ × 10¹² = 10¹⁵ pm

(ii) mg to kg, ng
👉 1 mg = 10^-3 g
👉 1 g = 10^-3 kg ⇒ 1 mg = 10^-3 × 10^-3 = 10^-6 kg
👉 1 mg = 10⁶ ng

(iii) mL to L, dm³
👉 1 mL = 10^-3 L
👉 1 L = 1 dm³ ⇒ 10^-3 L = 10^-3 dm³

Formula: Distance = Speed × Time

Given:
Speed of light, c = 3.00 × 10⁸ m s⁻¹
Time, t = 2.00 ns = 2.00 × 10⁻⁹ s

Distance = (3.00 × 10⁸) × (2.00 × 10⁻⁹) m
= 6.00 × 10⁻¹ m
= 0.600 m

Step 1: Understand the reaction: A + B₂ → AB₂
This means:
✅ 1 atom (or 1 mol) of A requires 1 molecule (or 1 mol) of B₂.
So ratio: A : B₂ = 1 : 1
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Step 2: Rule
👉 If A is less ⇒ A is limiting
👉 If B₂ is less ⇒ B₂ is limiting
👉 If both are equal ⇒ no limiting reagent
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(i) 300 atoms A + 200 molecules B₂
Need 300 B₂, available only 200 B₂
✅ B₂ is limiting (100 atoms A remain)
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(ii) 2 mol A + 3 mol B₂
Need 2 mol B₂, available 3 mol B₂
✅ A is limiting (1 mol B₂ remains)
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(iii) 100 atoms A + 100 molecules B₂
Required and available are equal
✅ No limiting reagent
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(iv) 5 mol A + 2.5 mol B₂
Need 5 mol B₂, available 2.5 mol B₂
✅ B₂ is limiting (2.5 mol A remains)
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(v) 2.5 mol A + 5 mol B₂
Need 2.5 mol B₂, available 5 mol B₂
✅ A is limiting (2.5 mol B₂ remains)
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Quick Trick
👉 Since ratio is 1:1, whichever is smaller is the limiting reagent.

Balanced equation: N₂ + 3H₂ → 2NH₃

Molar masses:
N₂ = 28 g mol⁻¹
H₂ = 2 g mol⁻¹
NH₃ = 17 g mol⁻¹

From stoichiometry:
28 g N₂ reacts with 6 g H₂ to form 34 g NH₃.

Step 1: H₂ required for 2.00 × 10³ g N₂
H₂ required = (6/28) × 2.00 × 10³
= 428.57 g

Available H₂ = 1.00 × 10³ g = 1000 g (excess)
✅ So N₂ is limiting, H₂ is excess.

Step 2: NH₃ formed from 2.00 × 10³ g N₂
NH₃ formed = (34/28) × 2.00 × 10³
= 2428.57 g ≈ 2.43 × 10³ g

Step 3: Unreacted H₂
H₂ consumed = 428.57 g
H₂ left = 1000 − 428.57 = 571.43 g

👉 Mole (mol): tells only “how much substance is present.”
Example: 0.50 mol Na₂CO₃ can be dissolved in 100 mL, 500 mL, or 2 L. Moles remain 0.50.

👉 Molarity (M): tells “moles per litre of solution.”
0.50 M means:
🤔 1 L solution contains 0.50 mol Na₂CO₃
🤔 2 L solution contains 1.00 mol Na₂CO₃

Reaction:
2H₂(g) + O₂(g) → 2H₂O(g)

By Gay-Lussac’s law of combining volumes (at same T and P), gaseous volumes combine in the ratio of coefficients.

Volume ratio from equation:
H₂ : O₂ : H₂O = 2 : 1 : 2

Given:
H₂ = 10 volumes, O₂ = 5 volumes
10 : 5 = 2 : 1 ✅ exact stoichiometric ratio

So both react completely.
Since 2 volumes H₂ give 2 volumes H₂O,
10 volumes H₂ will give 10 volumes H₂O.

(i) 28.7 pm to m
1 pm = 10^-12 m
28.7 pm = 28.7 × 10^-12 m = 2.87 × 10^-11 m

(ii) 15.15 pm to m
15.15 pm = 15.15 × 10^-12 m = 1.515 × 10^-11 m

(iii) 25365 mg to kg
1 mg = 10^-6 kg
25365 mg = 25365 × 10^-6 kg = 2.5365 × 10^-2 kg

📌 Number of atoms ∝ (moles × atoms per particle)
📌 Moles = mass / molar mass

Since each option has the same mass (1 g), the one with the lowest effective molar mass gives more particles. Also, if a molecule contains 2 atoms, total atoms increase accordingly.

(i) 1 g Au
Molar mass of Au ≈ 197 g mol⁻¹
Moles of atoms = 1/197 mol

(ii) 1 g Na
Molar mass of Na ≈ 23 g mol⁻¹
Moles of atoms = 1/23 mol

(iii) 1 g Li
Molar mass of Li ≈ 6.94 g mol⁻¹
Moles of atoms = 1/6.94 mol (highest among these)

(iv) 1 g Cl₂
Molar mass of Cl₂ ≈ 71 g mol⁻¹
Moles of molecules = 1/71 mol
Each molecule has 2 atoms, so moles of atoms = 2 × (1/71) = 2/71 mol

Comparison (in moles of atoms):
• Li: 1/6.94 ≈ 0.144
• Na: 1/23 ≈ 0.0435
• Au: 1/197 ≈ 0.0051
• Cl₂: 2/71 ≈ 0.0282

✅ The highest value is for Li, so 1 g Li contains the maximum number of atoms.
Note: For Cl₂, the factor of 2 is included because the question asks for number of atoms, not molecules.

Step 1: Formula for mole fraction
Mole fraction = moles of component / total moles

For ethanol:
0.040 = (moles of C₂H₅OH) / (moles of C₂H₅OH + moles of H₂O)

Let moles of ethanol = n
Then:
0.040 = n / (n + n(H₂O)) ……(i)

Step 2: Calculate moles of water in 1 L
Density of water = 1 g mL⁻¹
So, mass of 1 L water = 1000 g
Molar mass of water = 18 g mol⁻¹

n(H₂O) = 1000 / 18 = 55.55 mol

Substitute in (i):
0.040 = n / (n + 55.55)

Step 3: Solve for n (moles of ethanol)
0.040(n + 55.55) = n
0.040n + 2.222 = n
n − 0.040n = 2.222
0.960n = 2.222
n = 2.222 / 0.960 = 2.314 mol

Step 4: Molarity
Molarity = moles of solute / volume of solution in L
Molarity = 2.314 mol / 1 L = 2.314 mol L^-1

Step 1: Known fact
Molar mass of ¹²C = 12 g mol⁻¹
This means:
6.022 × 10²³ atoms of ¹²C have a mass of 12 g.

Step 2: Mass of one atom (unitary method)

Mass (g)	Number of atoms
12	6.022×1023
x	1

So,
x = 12 / (6.022×10²³) g
x ≈ 1.99 × 10⁻²³ g

(i) (0.02856 × 298.15 × 0.112) / 0.5785
This is multiplication/division.
Rule: final answer should have as many significant figures as the term with the least significant figures.

Number	Significant figures
0.02856	4
298.15	5
0.112	3
0.5785	4

Least = 3
✅ Final answer should have 3 significant figures.

(ii) 5 × 5.364
For this textbook-style question, 5 is generally treated as an exact counting factor, so it does not limit significant figures.
5.364 has 4 significant figures.
✅ Final answer: 4 significant figures.

(iii) 0.0125 + 0.7864 + 0.0215
This is addition.
Rule: final answer should have the same number of decimal places as the term with the least decimal places.

All terms have 4 decimal places.
✅ Final answer should be reported to 4 decimal places.

Formula:
Average molar mass = [Σ (abundance % × isotopic molar mass)] / 100

Substitute values:
M = [(0.337×35.96755) + (0.063×37.96272) + (99.600×39.9624)] / 100

Product terms:
0.337 × 35.96755 = 12.120 (approx)
0.063 × 37.96272 = 2.392 (approx)
99.600 × 39.9624 = 3980.255 (approx)

Sum = 12.120 + 2.392 + 3980.255 = 3994.767 (approx)

Divide by 100:
M = 3994.767 / 100 = 39.94767 g mol⁻¹

✅ Therefore, molar mass of natural argon ≈ 39.948 g mol⁻¹.

(i) 52 moles of Ar
1 mole contains Avogadro number of atoms:
N_A = 6.022 × 10²³ atoms mol⁻¹
So, atoms in 52 mol = 52 × 6.022 × 10²³
= 3.13144 × 10²⁵ atoms

(ii) 52 u of He
Atomic mass of He = 4 u per atom
Number of He atoms = 52 u ÷ 4 u = 13
✅ Therefore, 13 atoms

(iii) 52 g of He
Molar mass of He = 4 g mol⁻¹
Moles of He = 52 ÷ 4 = 13 mol
Number of atoms = 13 × 6.022 × 10²³
= 7.8286 × 10²⁴ atoms

Step 1: Calculate mass of carbon (C) from CO₂
Fraction of C in CO₂ = 12/44
Mass of C = (12/44) × 3.38 g = 0.9218 g

Step 2: Calculate mass of hydrogen (H) from H₂O
Fraction of H in H₂O = 2/18
Mass of H = (2/18) × 0.690 g = 0.0767 g

Since the fuel gas contains only C and H,
Total mass = 0.9218 + 0.0767 = 0.9985 g

(i) Calculation of Empirical Formula
Now calculate moles:
Moles of C = 0.9218/12 = 0.07682 mol
Moles of H = 0.0767/1 = 0.07670 mol

Simplest ratio:
C : H = 0.07682 : 0.07670 ≈ 1 : 1
✅ Therefore, empirical formula = CH


(ii) Molar Mass of the gas
At STP, 22.4 L gas = 1 mole
Given: mass of 10.0 L = 11.6 g
So, mass of 22.4 L:
(11.6/10.0) × 22.4 = 25.984 g

✅ Molar mass ≈ 26 g mol⁻¹ (approximately)


(iii) Molecular Formula
Empirical formula mass of CH = 12 + 1 = 13 g mol⁻¹

n = (Molar mass) / (Empirical formula mass)
n = 26/13 = 2

Molecular formula = (CH) × 2 = C₂H₂

Step 1: Calculate the amount of HCl in 25 mL of 0.75 M HCl

0.75 M HCl means that 1 liter (1000 mL) of solution contains 0.75 mol of HCl.

We know:
mass = moles × molar mass

Amount of HCl in 1000 mL:
= 0.75 mol × 36.5 g/mol = 27.37 g

Now, amount of HCl in 25 mL solution (Unitary method):

Volume (in mL)	Amount (in g)
1000	27.37
25	x

On solving for x:
x = (27.37 g × 25 mL) ÷ 1000 mL
x = 0.6845 g

Step 2: Stoichiometric ratio from the reaction
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

2 mol HCl (73 g) reacts with 1 mol CaCO₃ (100 g).

Step 3: Calculate mass of CaCO₃
For 73 g HCl, CaCO₃ required = 100 g
For 0.6845 g HCl, CaCO₃ required = (100/73) × 0.6845
= 0.938 g (approximately)

✅ Therefore, the mass of CaCO₃ required to react completely with 25 mL of 0.75 M HCl is 0.938 g.

Step 1: Find molar masses

Molar mass of HCl = (1 + 35.5) = 36.5 g mol^-1
So, mass of 4 mol HCl = 4 × 36.5 = 146 g

Molar mass of MnO₂ = 55 + (2 × 16) = 55 + 32 = 87 g mol^-1
So, mass of 1 mol MnO₂ = 87 g

Step 2: Use stoichiometric ratio from balanced equation
4HCl + MnO₂ → ...
Therefore, 1 mol MnO₂ reacts with 4 mol HCl

Hence, by mass:
✅ 87 g MnO₂ reacts with 146 g HCl

Step 3: Apply unitary method for 5.0 g MnO₂

MnO₂ (g)	HCl (g)
87	146
5.0	x

x = (146 × 5.0) / 87
x = 8.39 g

Reporting with correct significant figures (from 5.0 g):
✅ HCl required = 8.4 g